(Each question 1 marks)

Question 11 Mark

Write the sum of the series $1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + ..... + (2n - 1)^2 - (2n)^2$

Answer

We have, $1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + ..... + (2n - 1)^2 - (2n)^2 = (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + ...... + (2n - 1 - 2n)(2n - 1 + 2n) = -1[1 + 2 + 3 + 4 + ..... + 2n - 1 + 2n] =-1\Big[\frac{2\text{n}(2\text{n}+1)}{2}\Big]$ $\Big[\because\ 1 + 2\ ....\text{n} =\frac{\text{n}(\text{n}+1)}{2}\Big]$ $=-\text{n}(2\text{n}+1)$ Hence, sum of the series = -n(2n + 1)

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Question 21 Mark

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then write the value of k.

Answer

It is given that the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers. $\therefore\ 2+4+6+ \ ...\ +2\text{n}=\text{k}(1+3+5+\ ....\ +\text{n})$ $\Rightarrow2(1+2+3+ \ ...\ +\text{n})=\text{k}(1+3+5+\ ....\ +\text{n})$ $\Rightarrow2\times\frac{\text{n}(\text{n}+1)}{2}=\text{k}(1+3+5+\ ...\ +\text{n})$ $\Rightarrow\text{n}(\text{n}+1)=\text{k}(1+3+5+\ ...\ +\text{n})$ $\Rightarrow\text{n}(\text{n}+1)=\text{k}\times\text{n}^2$ $\big[\because$ sum of first odd natural numbers is $n^2 \big]$ $\Rightarrow\frac{(\text{n}+1)}{\text{n}}=\text{k}$ Hence, $\text{k}=\frac{\text{n}+1}{\text{n}}$

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Question 31 Mark

Let $S_n$ denote the sum of the cubes of first n natural numbers and $s_n$ denote the sum of first n natural numbers. Then, write the value of $\sum\limits^\text{n}_{\text{r}=1}\frac{\text{S}_\text{r}}{\text{S}_\text{r}}$

Answer

We know that, $\text{S}_\text{r}=1^3+2^3+3^3+\ ....\ +\text{r}^3=\Big[\frac{\text{r}(\text{r}+1)}{2}\Big]^2$ And, $\text{S}_\text{r}=1+2+3+\ ....\ +\text{r}=\frac{\text{r}(\text{r}+1)}{2}$ As, $\frac{\text{S}_\text{r}}{\text{S}_\text{r}}=\frac{\big[\frac{\text{r}(\text{r}+1)}{2}\big]^2}{\big[\frac{\text{r}(\text{r}+1)}{2}\big]}$ $=\frac{\text{r}(\text{r}+1)}{2}=\frac{1}{2}(\text{r}^2+\text{r})$ Now, $\sum\limits^{\text{n}}_{\text{r}=1}\frac{\text{S}_\text{r}}{\text{S}_\text{r}}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1}{2}(\text{r}^2+\text{r})$ $=\frac{1}{2}\Bigg(\sum\limits^{\text{n}}_{\text{r}=1}\text{r}^2+\sum\limits^{\text{n}}_{\text{r}=1}\text{r}\Bigg)$ $=\frac{1}{2}\Big[\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\frac{\text{n}(\text{n}+1)}{2}\Big]$ $=\frac{1}{2}\times\frac{\text{n}(\text{n}+1)}{2}\times\Big[\frac{(2\text{n}+1)}{3}+1\Big]$ $=\frac{\text{n}(\text{n}+1)}{4}\Big[\frac{2\text{n}+4}{3}\Big]$ $=\frac{\text{n}(\text{n}+1)}{4}\times\frac{2(\text{n}+2)}{3}$ $=\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$

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Question 41 Mark

If $\sum\limits^{\text{n}}_{\text{r}=1}\text{r}=55,$ find $\sum\limits^{\text{n}}_{\text{r}=1}\text{r}^3$

Answer

$\sum\limits^{\text{n}}_{\text{r}=1}\text{r}=55$ $\Rightarrow\frac{\text{n}(\text{n}+1)}{2}=55\ ...(\text{i})$ Now, $\sum\limits^{\text{n}}_{\text{r}=1}\text{r}^3=\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]^2$ $=\big[55\big]^2$ [Using equation (i), we get] $=3025$ Hence, $\sum\limits^{\text{n}}_{\text{r}=1}\text{r}^3=3025$

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Question 51 Mark

Write the sum of the series 2 + 4 + 6 + 8 + ... + 2n.

Answer

Let $T_n$ be the term of the given series and $S_n$ be the sum of the given series. $\therefore\ \text{T}_\text{n}=2\text{n}$ $\therefore\ \text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}=\sum\limits^{\text{n}}_{\text{k}=1}2\text{k}$ $=2\sum\limits^{\text{n}}_{\text{k}=1}\text{k}$ $=2\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]$ $=\text{n}(\text{n}+1)$ Hence, $\text{S}_\text{n}=\text{n}(\text{n}+1)$

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Question 61 Mark

Write the sum of 20 terms of the series $1+ \frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)+\ ...$

Answer

Let the $n^{th}$ term of the given series is $T_n$ and $S_n$ be the sum of the given series. $\therefore\ \text{T}_\text{n}=\frac{1}{\text{n}}\big[1+2+3+\ ...\ +\text{n}]$ $=\frac{1}{\text{n}}\Big[\frac{\text{n}}{2}\big(2\times1+(\text{n}-1)\times1\big)\Big]$ $=\frac{1}{\text{n}}\times\frac{\text{n}}{2}\big[2+\text{n}-1\big]$ $=\frac{1}{2}(\text{n}+1)$ $\therefore\ \text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\frac{1}{2}(\text{k}+1)$ $=\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}\text{k}+\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}1$ $=\frac{1}{2}\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]+\frac{1}{2}\times\text{n}$ $=\frac{\text{n}(\text{n}+1)}{4}+\frac{\text{n}}{2}$ $=\frac{\text{n}(\text{n}+1)+2\text{n}}{4}$ $=\frac{\text{n}[\text{n}+1+2]}{4}$ $=\frac{\text{n}[\text{n}+3]}{4}$ Putting, n = 20 we get $\text{S}_{20}=\frac{20[20+3]}{4}$ $=5\times23$ $=115$

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Question 71 Mark

Write the sum to n terms of a series whose $r^{th}$ term is $r + 2^r$

Answer

Let $T_n$ be the $n^{th}$ term of the series and $S_n$ be the sum to n terms of a series. $\therefore\ \text{T}_\text{r}=\text{r}+2^{\text{r}}$ [given] $\therefore\ \text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}(\text{r}+2^{\text{r}})$ $=\sum\limits^{\text{n}}_{\text{r}=1}\text{r}+\sum\limits^{\text{n}}_{\text{r}=1}2^{\text{r}}$ $=\frac{\text{n}(\text{n}+1)}{2}+\big[2^1+2^2\ ...\ 2^{\text{n}}\big]$ $=\frac{\text{n}(\text{n}+1)}{2}+\frac{2(2^\text{n}-1)}{(2-1)}$ $=\frac{\text{n}(\text{n}+1)}{2}+\frac{2\times2^\text{n}-2}{1}$ $=\frac{\text{n}(\text{n}+1)}{2}+2^{\text{n}+1}-2$ $\therefore\ \text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}+2^{\text{n}+1}-2$

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Question 81 Mark

Write the $50^{th}$ term of the series $2 + 3 + 6 + 11 + 18 + ....$

Answer

We have, $a_1=2, a_2=3=2+1, a_3=6=2+1+3, a_4=11=2+1+3+5$, ............................. ............................. .............................. $\mathrm{a}_{50}=2+1+3+5+\ldots .(50$ terms $)=2+\frac{49}{2}[2 \times 1+(49-1) \times 2]$ (As, the terms apart 2 are in A.P. with $\mathrm{a}=1$ and $\mathrm{d}=2)=2+\frac{49}{2}(2+48 \times 2)=2+\frac{49}{2} \times 98=2+49^2=2+2401=2403$ So, the $50^{\text {th }}$ term of the given series is 2403

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