Question

3+7+14+24+37+...

Answer 1

We have, 3+7+14+24+37+... The sequence of the differences between the successive terms of this series is 4,7,10,13+... clearly, it is an A.P. with common difference 3. Let $T_n$be the nth 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}+...(i)\\ Also, $S_n$ = 3 + 7 + 14 + 24 + 37 +...T_{n-1}+...(ii)\\ Subtracting (ii) from (i), we get 0=3+[4+7+10...$T_n-T_{n-1}]-T_n$ $\Rightarrow T_n=3+[4+7+\mathrm{10...}(T_n-T_{n-1}\left)\right]\Rightarrow T_n=3+\frac{(n+1)}{2}[2\times 4+(n-1-1)\times 3]=3+\frac{(n-1)}{2}[8+(n-2\left)3\right]=3+\frac{(n-1)}{2}[8+3n-6]=3+\frac{(n-1)}{2}[2+3n]=\frac{6+(n-1)(2+3n)}{2}$ $=\frac{6+2n+3n^2-2-3n}{2}=\frac{6+3n^2-n-2}{2}=\frac{3n^2-n+4}{2}\Rightarrow S_n={\sum }_{k-1}^h{T}_k={\sum }_{k-1}^h\frac{(3k^2-k+4)}{2}=\frac{1}{2}[{\sum }_{k-1}^{h}3k^2-{\sum }_{k-1}^h+{\sum }_{k-1}^{h}4]=\frac{3}{2}{\sum }_{k-1}^h{k}^2-\frac{1}{2}{\sum }_{k-1}^{h}k+{\sum }_{k-1}^{h}2$ $=\frac{3}{2}\left[\frac{n(n+1)(2n+1)}{6}\right]-\frac{1}{2}\left[\frac{n(n+1)}{2}\right]+2n=\frac{n(n+1)(2n+1)-n(n+1)+8n}{4}=\frac{n}{4}\left[\right(n+1\left)\right(2n+1)-(n+1)+8]=\frac{n}{4}[2n^2+n+2n+1-n-1+8]=\frac{n}{4}[2n^2+2n+8]=\frac{n}{4}[n^2+n+4]\times 2Hence,S_n=\frac{n}{2}[n^2+n+4]$