Solve the Following Question.(5 Marks) - Maths STD 11 Questions - Vidyadip

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Solve the Following Question.(5 Marks)

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22 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks

Find the sum of the following series to n terms:
$1.2.4 + 2.3.7 +3.4.10 + ...$

Answer

Let $T_n$ be the nth term of the givens series. Then,
$Tn = (n^{th}$ term of $1, 2, 3 ....) \times (n^{th}$ term of $2, 3, 4 ....) \times (n^{th}$ term of $4, 7, 10 ...)$
$= [1 + (n - 1) \times 1][2 + (n - 1) \times 1][4 + (n - 1) \times 3]$
$= [1 + n - 1][2 + n - 1][4 + 3n - 3]$
$= n(n + 1)(3n + 1)$
$= (n^2 + n)(3n + 1)$
$= 3n^3 + n^2 + 3n^2 + n$
$\therefore T_n = 3n^3 + 4n^2 + n$
Let $S_n$ denote the sum to n terms of the given series. Then,
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{n}=1}\text{T}_\text{n}$
$=\sum\limits^{\text{n}}_{\text{n}=1}(3\text{n}^3+4\text{n}^2+\text{n})$
$=\sum\limits^{\text{n}}_{\text{n}=1}3\text{n}^3+\sum\limits^{\text{n}}_{\text{n}=1}4\text{n}^2+\sum\limits^{\text{n}}_{\text{n}=1}\text{n}$
$=3\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]^2+4\Big[\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}\Big]+\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]$
$=\frac{3}{4}[\text{n}(\text{n}+1)]^2+\frac{2\text{n}(\text{n}+1)(2\text{n})+1}{3}+\frac{\text{n}(\text{n}+1)}{2}$
$=\frac{9\big[\text{n}(\text{n}+1)\big]^2+8\text{n}(\text{n}+1)(2\text{n}+1)+6\text{n}(\text{n}+1)}{12}$
$=\frac{\text{n}(\text{n}+1)}{12}\big[9\text{n}(\text{n}+1)+8(2\text{n}+1)+6\big]$
$=\frac{\text{n}}{12}(\text{n}+1)\big[9\text{n}^2+9\text{n}+16\text{n}+8+6\big]$
$=\frac{\text{n}}{12}(\text{n}+1)\big[9\text{n}^2+25\text{n}+14\big]$
Hence, $\text{S}_\text{n}=\frac{\text{n}}{12}(\text{n}+1)(9\text{n}^2+25\text{n}+14)$

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Question 25 Marks

Find the sum of the series whose $n^{th}$ term is:
$2n^3 + 3n^2 - 1$

Answer

We have,
$T_n = n^3 - 3n$
Let $S_n$ denote the sum of n terms of the series whose nth term is $T_n.$ Then,
Now,
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}(\text{k}^3-3^\text{k})$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^3-\sum\limits^{\text{n}}_{\text{k}=1}3\text{k}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^2(\text{n}+1)^2}{4}-(3+3^2+3^3+3^4+\ ....\ +3^{\text{n}})$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^2(\text{n}+1)^2}{4}-\Big[\frac{3(3^{\text{n}}-1)}{3-1}\Big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^2(\text{n}+1)^2}{4}-\frac{3}{2}(3^{\text{n}}-1)$

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Question 35 Marks

Sum the following series to $n$ terms:
$1 + 4 + 13 + 40 + 121 + .....$

Answer

We have,$1 + 4 + 13 + 40 + 121 + .....$
The sequence of the differences between the successive terms of the this series is $3, 9, 27, 81 ......$ Clearly, it is a G.P. with common difference $3.$
Let $T_n$ be the $n^{th}$ term and $S_n$ denote the sum of $n$ terms of the given series.
Then, $S_n = 1 + 4 + 13 + 40 + 121 + ..... + T_{n-1} + T_n ....(i)$
Also, $S_n = 1 + 4 + 13 + ...... + T_{n-1} + T_n ....(ii)$
Subtrating (ii) from (i), we get
$0=1+\big[3+9+27+81+\ ....\ +(\text{T}_\text{n}-\text{T}_{\text{n}-1})\big]-\text{T}_\text{n}$
$\text{T}_\text{n}=1+\big[3+9+27+81+\ ....\ +(\text{T}_\text{n}-\text{T}_{\text{n}-1})\big]$
$\text{T}_\text{n}=1+\frac{3\big(3^{\text{n}-1}-1\big)}{(3-1)}$
$\text{T}_\text{n}=1+\frac{3}{2}\big(3^{\text{n}-1}-1\big)$
$=1+\frac{3}{2}-3^{\text{n}-1}-\frac{3}{2}$
$=1-\frac{3}{2}+\frac{3^\text{n}}{2}$
$=-\frac{1}{2}+\frac{3^\text{n}}{2}$
$=\frac{3^\text{n}}{2}-\frac{1}{2}$
$\therefore\ \text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}=\sum\limits^{\text{n}}_{\text{k}=1}\Big(\frac{3^\text{n}}{2}-\frac{1}{2}\Big)$
$=\sum\limits^{\text{n}}_{\text{k}=1}\frac{3^\text{n}}{\text{n}}-\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}1$
$=\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}3^\text{n}-\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}1$
$\text{S}_\text{n}=\frac{1}{2}\big[3^1+3^2+3^3+\ .....\ +3^\text{n}\big]-\frac{1}{2}\times\text{n}$
$=\frac{1}{2}\Big[3\times\frac{(3^\text{n}-1)}{3-1}\Big]-\frac{\text{n}}{2}$
$=\frac{3\big(3^{\text{n}-1}\big)}{4}-\frac{\text{n}}{2}$
$=\frac{3\times3^\text{n}-3-2\text{n}}{4}$
$=\frac{3^{\text{n}+1}-2\text{n}-3}{4}$
Hence, $\text{S}_\text{n}=\frac{3^{\text{n}+1}-2\text{n}-3}{4}$

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Question 45 Marks

Sum the following series to $n$ terms:
$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ .....$

Answer

We have,
$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ .....$
Let $T_r$ be the rth term of the given series. Then,
$\text{T}_\text{r}=\frac{1}{(3\text{r}-2)(3\text{r}+1)},\text{r}=1,2,\ ...,\text{n}$
$\text{T}_\text{r}=\frac{1}{3}\Big[\frac{1}{3\text{r}-2}-\frac{1}{3\text{r}+1}\Big]$
$\therefore\ \text{required sum}=\frac{1}{3}=\sum\limits^{\text{n}}_{\text{r}=1}\text{T}_\text{r}$
$=\frac{1}{3}\sum\limits^{\text{n}}_{\text{r}=1}\Big[\frac{1}{3\text{n}-2}-\frac{1}{3\text{n}+1}\Big]$
$=\frac{1}{3}\Big[\Big(1-\frac{1}{4}\Big)+\Big(\frac{1}{4}-\frac{1}{7}\Big)+\Big(\frac{1}{7}-\frac{1}{100}\Big)\ ....\Big(\frac{1}{3\text{n}-2}-\frac{1}{3\text{n}+1}\Big)\Big]$
$=\frac{1}{3}\Big[1-\frac{1}{3\text{n}+1}\Big]$
$=\frac{1}{3}\Big[\frac{3\text{n}+1-1}{3\text{n}+1}\Big]$
$=\frac{1}{3}\times\frac{3\text{n}}{3\text{n}+1}$
Hence, required sum $=\frac{\text{n}}{3\text{n}+1}$

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Question 55 Marks

Find the sum of the following series to $n$ terms:
$1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ...$

Answer

Let $T_n$ be the nth term of the given series.
Thus, we have
$\text{T}_\text{n}=1+2+3+4+5+\ ...\ +\text{n}=\frac{\text{n}(\text{n}+1)}{2}=\frac{\text{n}^2+\text{n}}{2}$
Now, let $S_n$ be the sum of n terms of the given series.
Thus, we have
$\text{S}_\text{n}=\sum^\limits{\text{n}}_\text{k=1}\text{T}_\text{k}$
$\Rightarrow\text{S}_\text{n}=\sum^\limits{\text{n}}_\text{k=1}\Big(\frac{\text{k}^2+\text{k}}{2}\Big)$
$\Rightarrow\text{S}_\text{n}=\frac{1}{2}\sum\limits^{\text{n}}_{\text{k=1}}(\text{k}^2+\text{k})$
$\Rightarrow\text{S}_\text{n}=\frac{1}{2}\Big[\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\frac{\text{n}(\text{n}+1)}{2}\Big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{4}\Big(\frac{2\text{n}+1}{3}+1\Big)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{4}\Big(\frac{2\text{n}+4}{3}\Big)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)(2\text{n}+4)}{12}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$

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Question 65 Marks

Find the sum of the following series to n terms:
$1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...$

Answer

Let $T_n$ be the $n^{th}$ term of the given series. Then
$Tn = (n^{th}$ term of $1, 2, 3 .....) \times (n^{th}$ term of $2, 3, 4)$
$= [1 + (n + 1) \times 1][2 + (n + 1) \times 1]$
$= [1 + n - 1][2 + n - 1]$
$= n(n + 1)$
$= n^2 + n$
Let $S_n$ denote the sum to n terms of the given series. Then,
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{n}=1}\text{T}_\text{n}$
$=\sum\limits^{\text{n}}_{\text{n}=1}(\text{n}^2+\text{n})$
$=\sum\limits^{\text{n}}_{\text{n}=1}\text{n}^2+\sum\limits^{\text{n}}_{\text{n}=1}\text{n}$
$=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\frac{\text{n}(\text{n}+1)}{2}$
$=\frac{\text{n}(\text{n}+1)(2\text{n}+1)+3\text{n}(\text{n}+1)}{6}$
$=\frac{\text{n}(\text{n}+1)[2\text{n}+1+3]}{6}$
$=\frac{\text{n}(\text{n}+1)[2\text{n}+4]}{6}$
$=\frac{\text{n}(\text{n}+1)\times2(\text{n}+2)}{6}$
$=\frac{\text{n}}{6}(\text{n}+1)(\text{n}+2)$
Hence, $\text{S}_\text{n}=\frac{\text{n}}{3}(\text{n}+1)(\text{n}+2)$

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Question 75 Marks

Find the $20^{th}$ term and the sum of $20$ terms of the series:
$2 × 4 + 4 × 6 + 6 × 8 + ....$

Answer

Here the $n^{th}$ term of the series is:
$T_n = 2n(2n + 2)$
Thus the $20^{th}$ term will be,
$T_{20} = 2 \times 20(2 \times 20 + 2) = 1680$
The infinite series can be written as,
$2\times4+4\times6+6\times8+\ ....=\sum\limits^\infty_{\text{n}-1}2\text{n}(2\text{n}+2)$
Therefore the sum up to $20^{th}$ term will be,
$\sum\limits^{20}_{\text{n}=1}2\text{n}(2\text{n}+2)=\sum\limits^{20}_{\text{n}=1}4\text{n}^2+\sum\limits^{20}_{\text{n}=1}4\text{n}$
$=4\sum\limits^{20}_{\text{n}=1}\text{n}^2+4\sum\limits^{20}_{\text{n}=1}\text{n}$
$=4\times\frac{20(20+1)(2\times20+1)}{6}+4\times\frac{20(20+1)}{2}$
$=12320$

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Question 85 Marks

Find the sum of the series whose $n^{th}$ term is: $n(n + 1)(n + 4)$

Answer

$T_n = n(n + 1)(n + 4)$
$T_n = (n^2 + n)(n + 4)$
$T_n = n^3 + 5n^2 + 4n$
Let $S_n$ be the sum of n terms of the given series.
Now,
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}(\text{k}^3+5\text{k}^2+4\text{k})$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^3+5\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2+4\sum\limits^{\text{n}}_{\text{k}=1}\text{k}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^2(\text{n}+1)^2}{4}+\frac{5\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\frac{4\text{n}(\text{n}+1)}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}+\Big[\frac{\text{n}(\text{n}+1)}{2}+\frac{5(2\text{n}+1)}{3}+4\Big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\big[3\text{n}(\text{n}+1)+10(2\text{n}+1)+24\big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\big(3\text{n}^2+23\text{n}+34)$

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Question 95 Marks

Find the sum of the following series to $n$ terms:
$1^3 + 3^3 + 5^3 + 7^3 + ......$

Answer

Let $T_n$ be the $n^{th}$ term of the given series.
Thus, we have,
$\text{T}_\text{n}=(2\text{n}-1)^3$
Now, let $S_n$ be the sum of $n$ term of the given series.
Thus, we have,
$\text{S}_\text{n}=\sum\limits^\text{n}_{\text{k}=1}\text{T}_\text{k}$
$=\sum\limits^\text{n}_{\text{k}=1}\big[2\text{k}-1\big]^3$
$=\sum\limits^\text{n}_{\text{k}=1}\big[8\text{k}^3-1-6\text{k}(2\text{k}-1)\big]$
$=\sum\limits^\text{n}_{\text{k}=1}\big[8\text{k}^3-1-12\text{k}^2+6\text{k}\big]$
$=\sum\limits^\text{n}_{\text{k}=1}\big[8\text{k}^3-1-12\text{k}^2+6\text{k}\big]$
$=8\sum\limits^\text{n}_{\text{k}=1}\text{k}^3-\sum\limits^\text{n}_{\text{k}=1}1-12\sum\limits^\text{n}_{\text{k}=1}\text{k}^2+6\sum\limits^\text{n}_{\text{k}=1}\text{k}$
$=\frac{8\text{n}^2(\text{n}-1)}{4}-\text{n}-\frac{12\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\frac{6\text{n}(\text{n}+1)}{2}$
$=2\text{n}^2(\text{n}+1)^2-\text{n}-2\text{n}(\text{n}+1)(2\text{n}+1)+3\text{n}(\text{n}+1)$
$=\text{n}(\text{n}+1)-[2\text{n}(\text{n}+1)-2(2\text{n}+1)+3]-\text{n}$
$=\text{n}(\text{n}-1)\big[2\text{n}^2-2\text{n}+1\big]-\text{n}$
$=\text{n}\big[2\text{n}^3-2\text{n}^2+\text{n}+2\text{n}^2-2\text{n}+1-1\big]$
$=\text{n}\big[2\text{n}^3-\text{n}\big]$
$=\text{n}^2\big[2\text{n}^2-1\big]$

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Question 105 Marks

Find the sum of the following series to $n$ terms:
$1.2.5 + 2.3.6 + 3.4.7 + ...$

Answer

Let $T_n$ be the nth term of the given series.
Thus, we have
$T_n = n(n + 1)(n + 4)$
$T_n = n(n^2 + 5n + 4)$
$T_n = (n^3 + 5n^2 + 4n)$
Now, let Sn be the sum of n terms of the given series.
Thus, we have
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^3+\sum\limits^{\text{n}}_{\text{k}=1}5\text{k}^2+\sum\limits^{\text{n}}_{\text{k}=1}4\text{k}$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^3+5\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2+4\sum\limits^{\text{n}}_{\text{k}=1}\text{k}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^2(\text{n}+1)^2}{4}+\frac{5\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\frac{4\text{n}(\text{n}+1)}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^2(\text{n}+1)^2}{4}+\frac{5\text{n}(\text{n}+1)(2\text{n}+1)}{6}+2\text{n}(\text{n}+1)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big(\frac{\text{n}(\text{n}+1)}{2}+\frac{5(2\text{n}+1)}{3}+4\Big)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big(\frac{\text{n}^2+\text{n}}{2}+\frac{10\text{n}+5}{3}+4\Big)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big(\frac{3\text{n}^2+3\text{n}+20\text{n}+10+24}{6}\Big)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}(3\text{n}^2+23\text{n}+34)$

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Question 115 Marks

Find the sum of the following series to $n$ terms:
$2^2 + 4^2 + 6^2 + 8^2 + ....$

Answer

Let $T_n$ be the nth term of this series. Then,
$T_n = (2n)^3$
$T_n = 8n^3$
Let $S^n$ be the sum to n terms of the given series, Then,
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}8\text{k}^3$
$=8\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^3$
$=8\bigg[\frac{\text{n}(\text{n}+1)^2}{2}\bigg]^2$
$=8\times\frac{\text{n}^2(\text{n}+1)^2}{4}$
$=2\text{n}^2(\text{n}+1)^2$
Hence, $\text{S}_\text{n}=2\text{n}^2(\text{n}+1)^2$

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Question 125 Marks

Sum the following series to n terms:
$4 + 6 + 9 + 13 + 18 + .....$

Answer

We have,
$4 + 6 + 9 + 13 + 18 + .....$
The sequence of the differences between the successive terms of the this series is $2, 3, 4, 5 ......$ Clearly, it is a A.P. with common difference $1.$
Let $T_n$ be the $n^{th}$ term and $S_n$ denote the sum of $n$ terms of the given series.
Then, $S_n = 4 + 6 + 9 + 13 + 18 + ..... + T_{n-1} + T_n ....(i)$
Also, $S_n = 4 + 6 + 9 + 13 + ...... + T_{n-1} + T_n ....(ii)$
Subtrating (ii) from (i), we get
$0=4+\big[2+3+4+5+\ ....\ +(\text{T}_\text{n}-\text{T}_{\text{n}-1})\big]-\text{T}_\text{n}$
$\text{T}_\text{n}=4+\big[2+3+4+5\ +\ ....\ +(\text{T}_\text{n}-\text{T}_{\text{n}-1})\big]$
$\text{T}_\text{n}=4+\frac{(\text{n}-1)}{2}\big[2\times2+(\text{n}-1+1)\times1\big]$
$=4+\frac{\text{n}-1}{2}\big[4+\text{n}-2\big]$
$=4+\frac{(\text{n}-1)}{2}(\text{n}+2)$
$=\frac{8+\text{n}^2+2\text{n}-\text{n}-2}{2}$
$=\frac{\text{n}^2+\text{n}+6}{2}$
$\therefore\ \text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}=\sum\limits^{\text{n}}_{\text{k}=1}\Big(\frac{\text{k}^2+\text{k}+6}{2}\Big)$
$=\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2+\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}\text{k}+\sum\limits^{\text{n}}_{\text{k}=1}3$
$\text{S}_\text{n}=\frac{1}{2}\Big[\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}\Big]+\frac{\text{n}(\text{n}+1)}{2\times2}+3\text{n}$
$\text{S}_\text{n}=\frac{1}{12}(\text{n})(\text{n}+1)(2\text{n}+1)+\frac{\text{n}(\text{n}+1)}{4}+3\text{n}$
$=\frac{\text{n}(\text{n}+1)(2\text{n}+1)+3\text{n}(\text{n}+1)+36\text{n}}{12}$
$=\text{n}\Big[\frac{(\text{n}+1)(2\text{n}+1)+3(\text{n}+1)+36}{12}\Big]$
$=\frac{\text{n}}{12}\big[2\text{n}^2+\text{n}+2\text{n}+1+3\text{n}+3+36\big]$
$=\frac{\text{n}}{12}\big[2\text{n}^2+6\text{n}+40\big]$
$=\frac{2\text{n}}{12}\big[\text{n}^2+3\text{n}+20\big]$
$=\frac{\text{n}}{6}(\text{n}^2+3\text{n}+20)$
Hence, $\text{S}_\text{n}=\frac{\text{n}}{6}(\text{n}^2+3\text{n}+20)$

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Question 135 Marks

Sum the following series to n terms:
$2 + 5 + 10 + 17 + 26 + .....$

Answer

Let $T_n$ be the $n^{th}$ term and $S_n$ be the sum of n terms of the given series.
Thus, we have
$S_n = 2 + 5 + 10 + 17 + 26 + ..... + T_{n-1} + T_n ....(i)$
Equation (i) can be rewritten as,
$S_n = 2 + 5 + 10 + 17 + 26 + ...... + T_{n-1} + T_n ....(ii)$
On subtracting (ii) from (i), we get
$\underline{\underline{\text{S}_\text{n} = 2\ +\ 5\ +\ 10\ +\ 17 \ +\ 26\ +\ .........\ +\ \text{T}_{\text{n}-1}\ +\ \text{T}_\text{n}\\\text{S}_\text{n} =\ \ \ \ \ \ \ \ \ 2 \ +\ \ 5\ \ + \ 10\ +\ 17\ + 6 \ +\ ....\ \ +\ \text{T}_{\text{n}-1}\ +\ \text{T}_\text{n}}}\\\ 0\ =\ 2\ +\big[3\ +\ 5\ +\ 7\ +\ 9+\ .............\ +\ (\text{T}_\text{n}-\text{T}_{\text{n}-1})-\text{T}_\text{n}$
The sequence of difference of successive terms is $3, 5, 7, 9, .....$
We observe that it is an AP with common difference $2$ and first term $3.$
Thus, we have
$2+\Big[\frac{(\text{n}-1)}{2}\big\{6+(\text{n}-2)2\big\}\Big]-\text{T}_\text{n}=0$
$\Rightarrow2+\big[\text{n}^2+1\big]=\text{T}_\text{n}$
$\Rightarrow\big[\text{n}^2+1\big]=\text{T}_\text{n}$
Now,
$\because\ \text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}$
$\therefore\ \text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\big(\text{k}^2+1\big)$
$\Rightarrow\text{ S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2+\sum\limits^{\text{n}}_{\text{k}=1}1$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\text{n}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)(2\text{n}+1)+6\text{n}}{6}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(2\text{n}^2+3\text{n}+7)}{6}$

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Question 145 Marks

Sum the following series to $n$ terms:
$1 + 3 + 7 + 13 + 21 + .....$

Answer

We have,
$1 + 3 + 7 + 13 + 21 + .....$
The sequence of the differences between the successive terms of the this series is $2, 4, 6, 8 ......$ Clearly, it is an A.P. with common difference 2.
Let $T_n$ be the $n^{th}$ term and $S_n$ denote the sum of $n$ terms of the given series.
Then, $S_n = 1 + 3 + 7 + 13 + 21 + ...... + T_{n-1} + T_n .....(i)$
Also, $S_n = 1 + 3 + 7 + 13 + ...... + T_{n-1} + T_n ....(ii)$
Subtracting $(ii)$ from $(i),$ we get
$0=1+\big[2+4+6+8+\ ....\ +(\text{T}_\text{n}-\text{T}_{\text{n}-1})\big]-\text{T}_\text{n}$
$\text{T}_\text{n}=1+\big[2+4+6+8+\ ....\ +(\text{T}_\text{n}-\text{T}_{\text{n}-1})\big]$
$\text{T}_\text{n}=1+\frac{(\text{n}-1)}{2}\big[2\times2+(\text{n}-1-1)\times2\big]$
$\text{T}_\text{n}=1+\frac{(\text{n}-1)}{2}\times2\big[2+(\text{n}-2)\big]$
$\text{T}_\text{n}=1+(\text{n}-1)(\text{n})$
$\text{T}_\text{n}=1+\text{n}^2-\text{n}$
$\text{T}_\text{n}=\text{n}^2-\text{n}+1$
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}=\sum\limits^{\text{n}}_{\text{k}=1}\big(\text{k}^2-\text{k}+1\big)$
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2-\sum\limits^{\text{n}}_{\text{k}=1}\text{k}+\sum\limits^{\text{n}}_{\text{k}=1}1$
$=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}-\frac{\text{n}(\text{n}+1)}{2}+\text{n}$
$=\frac{\text{n}(\text{n}+1)(2\text{n}+1)-3\text{n}(\text{n}+1)+6\text{n}}{6}$
$=\frac{\text{n}}{6}\big[(\text{n}+1)(2\text{n}+1)-3(\text{n}+1)+6\big]$
$=\frac{\text{n}}{6}\big[2\text{n}^2+\text{n}+2\text{n}+1-3\text{n}-3+6\big]$
$=\frac{\text{n}}{6}\big[2\text{n}^2+4\big]$
$=\frac{\text{n}}{6}\times2\big[\text{n}^2+2\big]$
$=\frac{\text{n}}{3}(\text{n}^2+2)$
Hence, $\text{S}_\text{n}=\frac{\text{n}}{3}(\text{n}^2+2)$

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Question 155 Marks

Find the sum of the series whose $n^{th}$ term is:$(2n - 1)^2$

Answer

$T_n = (2n - 1)^2$
Let $S_n$ be the sum of $n$ terms of the given series.
Now,
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}(2\text{k}-1)^2$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}(4\text{k}^2+1-4\text{k})$
$\Rightarrow\text{S}_\text{n}=4\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2+\sum\limits^{\text{n}}_{\text{k}=1}1-4\sum\limits^{\text{n}}_{\text{k}=1}\text{k}$
$\Rightarrow\text{S}_\text{n}=\frac{4\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\text{n}-\frac{4\text{n}(\text{n}+1)}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big[\frac{4(2\text{n}+1)}{3}-4\Big]+\text{n}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big(\frac{8\text{n}+4-12}{3}\Big)+\text{n}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big(\frac{8\text{n}-8}{3}\Big)+\text{n}$
$\Rightarrow\text{S}_\text{n}=4\text{n}(\text{n}+1)\Big(\frac{\text{n}-1}{3}\Big)+\text{n}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}(4\text{n}^2+4\text{n}-4\text{n}-4+3)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}(4\text{n}^2-1)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}(2\text{n}-1)(2\text{n}+1)$

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Question 165 Marks

Sum the following series to n terms:
$3 + 7 + 14 + 24 + 37 + .....$

Answer

We have,
$3 + 7 + 14 + 24 + 37 + .....$
The sequence of the differences between the successive terms of the this series is $4, 7, 10, 13 + ....$ Clearly, it is an A.P. with common difference $3$.
Let $T_n$ be the $n^{th}$ term and $S_n$ denote the sum of n terms of the given series.
Then, $S_n = 3 + 7 + 14 + 24 + 37 + ..... + T_{n-1} + T_n ....(i)$
Also, $S_n = 3 + 7 + 14 + 24 + ...... + T_{n-1} + T_n ....(ii)$
Subtrating (ii) from (i), we get
$0=3+\big[4+7+10+\ ....\ +(\text{T}_\text{n}-\text{T}_{\text{n}-1})\big]-\text{T}_\text{n}$
$\text{T}_\text{n}=3+\big[4+7+10+\ ....\ +(\text{T}_\text{n}-\text{T}_{\text{n}-1})\big]$
$\text{T}_\text{n}=3+\frac{(\text{n}-1)}{2}\big[2\times4+(\text{n}-1-1)\times3\big]$
$=3+\frac{(\text{n}-1)}{2}\big[8+(\text{n}-2)3\big]$
$=3+\frac{(\text{n}-1)}{2}\big[8+3\text{n}-6\big]$
$=3+\frac{(\text{n}-1)}{2}\big[2+3\text{n}\big]$
$=\frac{6+(\text{n}-1)(2+3\text{n})}{2}$
$=\frac{6+2\text{n}+3\text{n}^2-2-3\text{n}}{2}$
$=\frac{6+3\text{n}^2-\text{n}-2}{2}$
$=\frac{3\text{n}^2-\text{n}-2}{2}$
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}=\sum\limits^{\text{n}}_{\text{k}=1}\frac{\big(3\text{k}^2-\text{k}+4\big)}{2}$
$=\frac{1}{2}\Bigg[\sum\limits^{\text{n}}_{\text{k}=1}3\text{k}^2-\sum\limits^{\text{n}}_{\text{k}=1}\text{k}+\sum\limits^{\text{n}}_{\text{k}=1}4\Bigg]$
$=\frac{3}{2}\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2-\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}\text{k}+\sum\limits^{\text{n}}_{\text{k}=1}2$
$=\frac{3}{2}\Big[\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}\Big]-\frac{1}{2}\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]+2\text{n}$
$=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{4}-\frac{\text{n}(\text{n}+1)}{4}+2\text{n}$
$=\frac{\text{n}(\text{n}-1)(2\text{n}+1)-\text{n}(\text{n}+1)+8\text{n}}{4}$
$=\frac{\text{n}}{4}\big[(\text{n}+1)(2\text{n}+1)-(\text{n}+1)+8\big]$
$=\frac{\text{n}}{4}\big[2\text{n}^2+\text{n}+2\text{n}+1-\text{n}-1+8\big]$
$=\frac{\text{n}}{4}\big[2\text{n}^2+2\text{n}+8\big]$
$=\frac{\text{n}}{4}\times2\big[\text{n}^2+\text{n}+4\big]$
$=\frac{\text{n}}{2}\big[\text{n}^2+\text{n}+4\big]$
Hence, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[\text{n}^2+\text{n}+4\big]$

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Question 175 Marks

Sum the following series to n terms:
$3 + 5 + 9 + 15 + 23 + .....$

Answer

We have,
$3 + 5 + 9 + 15 + 23 + ..... + T_{n-1} + T_n$
The difference between the successive terms are $5 - 3 = 2, 9 - 5 = 4, 15 - 9 = 6 ..... $ Clearly, these difference are in A.P.
Let, S n denote the sum to n terms of the given series.
Then,
$S_n = 3 + 5 + 9 + 15 + 23 + ..... + T_{n-1} + T_n ....(i)$
Also, $S_n = 3 + 5 + 9 + 15 + ..... + T_{n-1} + T_n ....(ii)$
Subtrating (ii) from (i), we get
$0=3+\big[2+4+6+8+\ ....\ +(\text{T}_\text{n}-\text{T}_{\text{n}-1})\big]-\text{T}_\text{n}$
$\text{T}_\text{n}=3+\frac{(\text{n}-1)}{2}\big[2\times2+(\text{n}-1-1)\times2\big]$
$\text{T}_\text{n}=3+\frac{(\text{n}-1)}{2}\times2\big[2+\text{n}-2\big]$
$\text{T}_\text{n}=3+(\text{n}-1){\text{n}}$
$\text{T}_\text{n}=3+\text{n}^2-\text{n}$
$\text{T}_\text{n}=\text{n}^2-\text{n}+3$
$\therefore\ \text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}=\sum\limits^{\text{n}}_{\text{k}=1}\big(\text{k}^2-\text{k}+3\big)$
$=\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2-\sum\limits^{\text{n}}_{\text{k}=1}\text{k}+\sum\limits^{\text{n}}_{\text{k}=1}3$
$=\frac{\text{n}(\text{n}+1)(2\text{n}+1)-\text{n}(\text{n}+1)}{6}+3\text{n}$
$=\frac{\text{n}(\text{n}+1)(2\text{n}+1)-\text{n}(\text{n}+1)+18\text{n}}{6}$
$=\frac{\text{n}}{6}\big[(\text{n}+1)(2\text{n}+1)-3(\text{n}+1)+18\big]$
$=\frac{\text{n}}{6}\big[2\text{n}^2+\text{n}+2\text{n}+1-3\text{n}-3+18\big]$
$=\frac{\text{n}}{6}\big[2\text{n}^2+3\text{n}-3\text{n}-2+18\big]$
$=\frac{\text{n}}{6}\big[2\text{n}^2+16\big]$
$=\frac{\text{n}}{6}\times2\big[\text{n}^2+8\big]$
$=\frac{\text{n}}{3}(\text{n}^2+8)$
Hence, $\text{S}_\text{n}=\frac{\text{n}}{3}\big(\text{n}^2+8\big)$

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Question 185 Marks

Sum the following series to n terms:
$\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+\ .....\ +\frac{1}{(5\text{n}-4)(5\text{n}+1)}$

Answer

We have,
$\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+\ .....\ +\frac{1}{(5\text{n}-4)(5\text{n}+1)}$
Let $T_r$ be the rth term of the given series. Then,
$\text{T}_\text{r}=\frac{1}{(5\text{r}-4)(5\text{r}+1)},\text{r}=1,2,\ ...,\text{n}$
$\text{T}_\text{r}=\frac{1}{5}\Big[\frac{1}{5\text{r}-2}-\frac{1}{5\text{r}+1}\Big]$
$\therefore\ \text{required sum}=\frac{1}{5}=\sum\limits^{\text{n}}_{\text{r}=1}\text{T}_\text{r}$
$=\frac{1}{5}\sum\limits^{\text{n}}_{\text{r}=1}\Big[\frac{1}{5\text{r}-4}-\frac{1}{5\text{r}+1}\Big]$
$=\frac{1}{5}\Big[\Big(1-\frac{1}{6}\Big)+\Big(\frac{1}{6}-\frac{1}{11}\Big)+\Big(\frac{1}{11}-\frac{1}{14}\Big)\ ....\Big(\frac{1}{5\text{n}-4}-\frac{1}{5\text{n}+1}\Big)\Big]$
$=\frac{1}{5}\Big[1-\frac{1}{5\text{n}+1}\Big]$
$=\frac{1}{5}\Big[\frac{5\text{n}+1-1}{5\text{n}+1}\Big]$
$=\frac{1}{5}\times\frac{5\text{n}}{5\text{n}+1}$
$=\frac{\text{n}}{5\text{n}+1}$
Hence, required sum $=\frac{\text{n}}{5\text{n}+1}$

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Question 195 Marks

Sum the following series to $n$ terms:
$2 + 4 + 7 + 11 + 16 + ......$

Answer

We have,
$2 + 4 + 7 + 11 + 16 + ......$
The sequence of the differences between the successive terms of the this series is $2, 3, 4, 5 ......$ Clearly, it is a A.P. with common difference 1.
Let $T_n $be the $n^{th}$ term and $S_n$ denote the sum of $n$ terms of the given series.
Then, $S_n = 2 + 4 + 7 + 11 + 16 + ....... + T_{n-1} + T_n ....(i)$
Also, $S_n = 2 + 4 + 7 + 11 + ...... + T_{n-1} + T_n ....(ii)$
Subtrating (ii) from (i), we get
$0=2+\big[2+3+4+5+\ ....\ +(\text{T}_\text{n}-\text{T}_{\text{n}-1})\big]-\text{T}_\text{n}$
$\text{T}_\text{n}=2+\frac{(\text{n}-1)}{2}\big[2\times2+(\text{n}-1-1)\times1\big]$
$=2+\frac{(\text{n}-1)}{2}\big[4+\text{n}-2\big]$
$=2+\frac{(\text{n}-1)}{2}(\text{n}+2)$
$=\frac{4+\text{n}^2+2\text{n}-\text{n}-2}{2}$
$=\frac{\text{n}^2+\text{n}+2}{2}$
$=\frac{\text{n}^2}{2}+\frac{\text{n}}{2}+1$
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}=\sum\limits^{\text{n}}_{\text{k}=1}\Big(\frac{\text{k}^2}{2}+\frac{\text{k}}{2}+1\Big)$
$=\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}\text{n}^2+\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}\text{n}+\sum\limits^{\text{n}}_{\text{k}=1}1$
$=\frac{1}{2}\Big[\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}\Big]+\frac{1}{2}\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]+\text{n}$
$\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{12}+\frac{\text{n}(\text{n}+1)}{4}+\text{n}$
$=\frac{\text{n}(\text{n}+1)(2\text{n}+1)+3\text{n}(\text{n}+1)+12\text{n}}{12}$
$=\frac{\text{n}}{12}\big[(\text{n}+1)(2\text{n}+1)+3(\text{n}+1)+12\big]$
$=\frac{\text{n}}{12}\big[2\text{n}^2+6\text{n}+16\big]$
$=\frac{2\text{n}}{12}\big[\text{n}^2+3\text{n}+8\big]$
$=\frac{\text{n}}{6}\big[\text{n}^2+3\text{n}+8\big]$
Hence, $\text{S}_\text{n}=\frac{\text{n}}{6}\big(\text{n}^2+3\text{n}+8\big)$

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Question 205 Marks

Find the sum of the following series to n terms:
$3 \times 1^2 + 5 \times 2^2 + 7 \times 3^2 + ...$

Answer

Let $T_n$ the $n^{th}$ term of the given series. Then,
$T_n$ = ($n^{th}$ term of $3, 5, 7 ....$) $\times $ ($n^{th}$ term of $1^2, 2^2, 3^2 ....$)
$= [3 + (n - 1)2][n^2]$
$= [3 + 2n - 2][n^2]$
$=[2n + 1][n^2]$
$= 2n^3 + n^2$
$\therefore$ $T_n = 2n^3 + n^2$
Let $S_n$ denote the sum of n terms of the given series. Then,
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{n}=1}\text{T}_\text{n}$
$=\sum\limits^{\text{n}}_{\text{n}=1}(2\text{n}^3+\text{n}^2)$
$=\sum\limits^{\text{n}}_{\text{n}=1}2\text{n}^3+\sum\limits^{\text{n}}_{\text{n}=1}\text{n}^2=2\sum\limits^{\text{n}}_{\text{n}=1}\text{n}^3+\sum\limits^{\text{n}}_{\text{n}=1}\text{n}^2$
$=2\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]^2+\Big[\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}\Big]$
$=\frac{2}{4}\big[\text{n}(\text{n}+1)\big]^2+\frac{\big[\text{n}(\text{n}+1)(2\text{n}+1)\big]}{6}$
$=\frac{[\text{n}(\text{n}+1)]^2}{2}+\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$=\frac{3[\text{n}(\text{n}+1)]^2+\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$=\frac{\text{n}(\text{n}+1)}{6}\big[3\text{n}(\text{n}+1)+(2\text{n}+1)\big]$
$=\frac{\text{n}(\text{n}+1)}{6}\big[3\text{n}^2+3\text{n}+2\text{n}+1\big]$
$=\frac{\text{n}}{6}(\text{n}+1)(3\text{n}^2+5\text{n}+1)$
Hence, $\text{S}_\text{n}=\frac{\text{n}}{6}(\text{n}+1)(3\text{n}^2+5\text{n}+1)$

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Question 215 Marks

Sum the following series to n terms:
$1 + 3 + 6 + 10 + 15 + .....$

Answer

We have,
$1 + 3 + 6 + 10 + 15 + .....$
The sequence of the differences between the successive terms of the this series is 2, 3, 4, 5 + .... Clearly, it is an A.P. with common difference 1.
Let $T_n$ be the $n^{th}$ term and $S_n$ denote the sum of n terms of the given series.
Then, $S_n = 1 + 3 + 6 + 10 + 15 + ..... + T_{n-1} + T_n ....(i)$
Also, $S_n = 1 + 3 + 6 + 10 + 15 + ...... + T_{n-1} + T_n ....(ii)$
Subtrating (ii) from (i), we get
$0=1+\big[2+3+4+5+\ ....\ +(\text{T}_\text{n}-\text{T}_{\text{n}-1})\big]-\text{T}_\text{n}$
$\text{T}_\text{n}=1+\big[2+3+4+5+\ ....\ +(\text{T}_\text{n}-\text{T}_{\text{n}-1})\big]$
$\text{T}_\text{n}=1+\frac{(\text{n}-1)}{2}\big[2\times2+(\text{n}-1-1)\times1\big]$
$=1+\frac{(\text{n}-1)}{2}\big[4+\text{n}-2\big]$
$=1+\frac{(\text{n}-1)}{2}(\text{n}+2)$
$=1+\frac{\text{n}^2+2\text{n}-\text{n}-2}{2}$
$=1+\frac{\text{n}^2+\text{n}-2}{2}$
$=\frac{2+\text{n}^2+\text{n}-2}{2}$
$=\frac{\text{n}^2+\text{n}}{2}$
$\therefore\ =\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}=\sum\limits^{\text{n}}_{\text{k}=1}\Big(\frac{\text{k}^2+\text{k}}{2}\Big)$
$=\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2+\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}\text{k}$
$\Rightarrow\text{S}_\text{n}=\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2+\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}\text{k}$
$=\frac{1}{2}\Big[\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\frac{\text{n}(\text{n}+1)}{2}\Big]$
$=\frac{1}{2}\bigg[\text{n}(\text{n}+1)\Big[\frac{2\text{n}+1}{6}+\frac{1}{2}\Big]\bigg]$
$=\frac{\text{n}(\text{n}+1)}{2}\Big[\frac{2\text{n}+1+3}{6}\Big]$
$=\frac{\text{n}(\text{n}+1)}{2}\Big[\frac{2\text{n}+4}{6}\Big]$
$=\frac{\text{n}(\text{n}+1)\times2[\text{n}+2]}{2\times6}$
$=\frac{\text{n}}{6}(\text{n}+1)(\text{n}+2)$
Hence, $\text{S}_\text{n}=\frac{\text{n}}{6}(\text{n}+1)(\text{n}+2)$

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Question 225 Marks

Find the sum of the series whose $n^{th}$ term is:
$2n^3 + 3n^2 - 1$

Answer

$T_n = 2n^3 + 3n^2 - 1$ Let $S_n $ be the sum of n terms of the given series.
Now,
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}(2\text{k}^2+3\text{k}-1)$
$\Rightarrow\text{S}_\text{n}=2\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2+3\sum\limits^{\text{n}}_{\text{k}=1}\text{k}-\sum\limits^{\text{n}}_{\text{k}=1}1$
$\Rightarrow\text{S}_\text{n}=\frac{2\text{n}^2(\text{n}+1)^2}{4}+\frac{3\text{n}(\text{n}+1)(2\text{n}+1)}{6}-\text{n}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^2(\text{n}+1)^2}{2}+\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{2}-\text{n}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^2(\text{n}+1)^2+\text{n}(\text{n}+1)(2\text{n}+1)-2\text{n}}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^2(\text{n}^2+1+2\text{n})+(\text{n}^2+\text{n})(2\text{n}+1)-2\text{n}}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^2(\text{n}^2+1+2\text{n})+(2\text{n}^3+\text{n}^2+2\text{n}^2+\text{n})-2\text{n}}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}^4+4\text{n}^2+4\text{n}^3-\text{n}}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}^3+4\text{n}+4\text{n}^2-1)}{2}$

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