The sum of n terms of the series 1+5+12+22+35+ is

Correct Answer is : n2(n+1)2\;n²(n+1)/22n2(n+1)

Correct Answer is : n2(n+1)2\;n²(n+1)/22n2(n+1)

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CGPET 2014 Solved Paper

Section: Mathematics

Question : 135 of 150

Marks: +1, -0

The sum of

n

1+5+12+22+35+\dots

Solution:

- Let

\;S_n=1+5+12+22+35+\dots+T_n

\;S_n=1+5+12+22+\dots+T_n

\;0=1+4+7+10+13+\dots-T_n

\;\Rightarrow T_n=1+4+7+10+13+\dots+n\;\text{ terms }\;

This is an arithmetic series whose first term a is 1 and common difference

d

is 3 .

\therefore\;T_n=\;\frac{n}{2}[2 \times 1+(n-1) 3]

\Rightarrow T_n=\;\frac{n}{2}[2+3 n-3]

\;=\;\frac{n}{2}[3 n-1]

\;=\;\frac{1}{2}[3 n^2-n]

S_n\;=\sum T_n

\;=\;\frac{1}{2}\left[\sum (3 n^2-n)\right]

\;=\;\frac{1}{2}\left[3\sum n^2 - \sum n\right]

\;=\;\frac{1}{2}\left[\;\frac{3 n(n+1)(2 n+1)}{6} - \;\frac{n(n+1)}{2}\right]

\;=\;\frac{n(n+1)}{2}\left[\;\frac{2 n+1}{2} - \;\frac{1}{2}\right]

\;=\;\frac{n(n+1)}{2}\left[\;\frac{2 n}{2}\right]=\;\frac{n^2(n+1)}{2}

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