Question

4 + 6 + 9 + 13 + 18 +....

Answer 1

We have, 4 + 6 + 9 + 13 + 18 +.... The sequence of the differences between the successive terms of this series is 2, 3, 4, 5+... clearly, it is an A.P. with common difference 1. Let $T_n$ be the nth term and $S_n$ denote the sum of n terms of the given series. Then, $S_n=4+6+13+\mathrm{18...}{T}_{n-1}+T_n...\left(i\right)$ Also, $S_n=4+6+13+\mathrm{18...}{T}_{n-1}+T_n...\left(i\right)$ Subtracting (ii) from (i),we get $0=4+[2+3+4+\mathrm{5...}(T_n-T_{n-1}\left)\right]-T_n\Rightarrow T_n=4+[2+3+4+\mathrm{5...}(T_n-T_{n-1}\left)\right]\Rightarrow T_n=4+\frac{(n-1)}{2}[2\times 2+(n-1-1)\times 1]=4+\frac{(n-1)}{2}[4+n-2]=4+\frac{(n-1)}{2}(n+2)=\frac{8+n^2+2n-n-2}{2}=\frac{n^2+n+6}{2}\Rightarrow S_n={\sum }_{k-1}^n{T}_k={\sum }_{k-1}^n\left(\frac{k^2+k+6}{2}\right)=\frac{1}{2}{\sum }_{k-1}^n{k}^2+\frac{1}{2}{\sum }_{k-1}^{n}k+{\sum }_{k-1}^{n}3\Rightarrow S_n=\frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6}\right]+\frac{n(n+1)}{2\times 2}+3n{S}_n=\frac{1}{12}\left(n\right)(n+1)(2n+1)+\frac{n(n+1)}{4}+3n=\frac{n(n+1)(2n+1)+3(n+1)+36n}{12}=n\left[\frac{(n+1)(2n+1)+3(n+1)+36}{12}\right]=\frac{n}{12}[2n^2+n+2n+1+3n+3+36]=\frac{n}{12}[2n^2+6n+40]=\frac{2n}{12}[n^2+3n+20]=\frac{n}{6}[n^2+3n+20]Hence,S_n=\frac{n}{6}[n^2+3n+20]$