Question

Find the sum of he first n terms of the series $3+7+12+21+31+\cdots$

Answer 1

Let $S_n=3+7+13+21+31+\cdots +T_{n-1}+T_n{S}_n=3+7+13+21+\cdots +T_{n-2}+T_{n-1}+T_n\text{On Subtraction, we get}0=3+[4+6+8+10\cdots \text{To}(n-1\left)\text{terms}\right]=3+\frac{(n-1)}{2}\times [2\times 4+(n-2)\times 2]=3+(n-1)(n+2)=(n^2+n+1).\therefore S_n={\sum }_{k=1}^n{T}_k={\sum }_{k=1}^n(k^2+k+1)={\sum }_{k=1}^n{k}^2+{\sum }_{k=1}^{n}k+n=\frac{1}{6}n(n+1)(2n+1)+\frac{1}{2}n(n+1)+n=\frac{1}{3}n(n^2+3n+5).$