Question

Find the sum of the following series to n terms 1.2.4+2.3.7+3.4.10+.....

Answer 1

Let $T_n$be the nth term of this series. Then, $T_n=(nthtermof1,2,\mathrm{3...})\times (nthtermof2,3,\mathrm{4...})\times (nthtermof4,7,\mathrm{10...})=[1+(n-1)\times 1].[2+(n-1)\times 1].[4+(n-1)\times 3]=[1+n-1].[2+n-1].[4+3n-3]=n(n+1)(3n+1)=(n^2+n)(3n+1)=3n^3+n^2+3n^2+n=3n^3+4n^2+n\therefore T_n=3n^3+4n^2+nLet{S}_n\text{be the sum to n terms of the given series; Then,}{S}_n={\sum }_{n-1}^n{T}_n={\sum }_{n-1}^n(3n^3+4n^2+n)={\sum }_{n-1}^{n}3n^3+{\sum }_{n-1}^{n}4n^2+{\sum }_{n-1}^{n}n=3{\sum }_{n-1}^n{n}^3+4{\sum }_{n-1}^n{n}^2+{\sum }_{n-1}^{n}n={\left[\frac{n(n+1)}{2}\right]}^2+4\left[\frac{n(n+1)(2n+1)}{2}\right]+\left[\frac{n(n+1)}{2}\right]=\frac{3}{4}\left[n\right(n+1){]}^2+\frac{2n(n+1)(2n+1)}{3}+\frac{n(n+1)}{2}=\frac{9\left[n\right(n+1){]}^2+8n(n+1\left)\right(2n+1)+6n(n+1)}{12}=\frac{n(n+1)}{12}[9n(n+1)+8(2n+1)+6]=\frac{n}{12}(n+1\left)\right[9n^2+9n+16n+8+6]=\frac{n}{12}(n+1\left)\right[9n^2+25n+14]Hence,S_n=\frac{n}{12}(n+1\left)\right(9n^2+25n+14)$