Find the sum of n terms of the series 3 + 7 + 14 + 24 + 37 + …….

Solution :

By using method of differences,

The difference between the successive terms are 7 – 3 = 4, 14 – 7 = 7, 24 – 14 = 10, …. Clearly,these differences are in AP.

Let $T_n$ be the nth term and $S_n$ denote the sum to 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 + 37 + ….. + $T_{n-1}$ + $T_n$ ……(ii)

Subtracting (ii) from (i), we get

0 = 3 + [4 + 7 + 10 + 13 + …… + $T_{n-1}$ + $T_n$] – $T_n$

$\implies$ $T_n$ = 3 + $(n-1)\over 2${2*4+(n-1-1)*3} = $6 + (n-1)(3n+2)\over 4$

$\implies$ $T_n$ = $1\over 2$ ($3n^2 – n + 4$)

$\therefore$ $S_n$ = $\sum_{k=1}^{n}$ $T_k$ = $\sum_{k=1}^{n}$ $1\over 2$ ($3n^2 – n + 4$) = $1\over 2$[$\sum_{k=1}^{n}$$3k^2$ – $\sum_{k=1}^{n}$ k + 4n]

$\implies$ $S_n$ = $1\over 2$[3$n(n+1)(2n+1)\over 6$  – $n(n+1)\over 2$ + 4n] = $n\over 2$ ($n^2 + n + 4$)

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