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Some Special Series — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSome Special Series5 Marks
Question
Sum the following series to n terms:
$3 + 5 + 9 + 15 + 23 + .....$
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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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