Question

Sum the following series to n terms : 3+5+9+15+23+....

Answer 1

We have, 3+5+9+15+23+....+$T_{n-1}+T_n{S}_n=3+5+9+15+23+T_{n-1}+T_n...\left(i\right)Also,S_n=3+5+9+\mathrm{15...}{T}_{n-1}+T_n...\left(i\right)Subtracting\left(ii\right)from\left(i\right),weget0=3+[2+4+6+\mathrm{8...}(T_n-T_{n-1}\left)\right]-Tn{T}_n=3+\frac{(n-1)}{2}[2\times 2+(n-1-1)\times 2]T_n=3+\frac{(n-1)}{2}\times 2[2+n-2]=3+(n-1)\left(n\right)=3+n^2-n=n^2-n+3\therefore S_n={\sum }_{k-1}^h{T}_k={\sum }_{k-1}^h(k^2-k+3)={\sum }_{k-1}^h{k}^2-{\sum }_{k-1}^{h}k+{\sum }_{k-1}^{h}3=\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+3n=\frac{n(n+1)(2n+1)-3n(n+1)+18n}{6}=\frac{n}{6}\left[\right(n+1\left)\right(2n+1)-3(n+1)+18]=\frac{n}{6}[2n^2+n+2n+1-3n-3+18]=\frac{n}{6}[2n^2+3n-3n-2+18]=\frac{n}{6}[2n^2+16]=\frac{n}{6}\times 2[n^2+8]=\frac{n}{3}(n^2+8)Hence,S_n=\frac{n}{2}(n^2+8)$