Some Special Series — MATHS STD 11 Science — Question
Gujarat BoardEnglish MediumSTD 11 ScienceMATHSSome Special Series4 Marks
Question
Sum the following series to n terms: $3 + 7 + 14 + 24 + 37 + .....$
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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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