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Some Special Series — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSSome Special Series5 Marks
Question
Sum the following series to n terms:
2 + 4 + 7 + 11 + 16 + ......
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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 Tn be the nth term and Sn denote the sum of n terms of the given series.
Then, Sn = 2 + 4 + 7 + 11 + 16 + ....... + Tn-1 + Tn ....(i)
Also, Sn = 2 + 4 + 7 + 11 + ...... + Tn-1 + Tn ....(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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