Question

$1.25+\mathrm{2.3.6}+\mathrm{3.4.7}+....$

Answer 1

$1.25+\mathrm{2.3.6}+\mathrm{3.4.7}+....Let{T}_n\text{be the nth term of this series.}Then,T_n=(nthtermof1,2,\mathrm{3...})\times (nthtermof2,3,\mathrm{4...})\times (nthtermof5,6,\mathrm{7...})=[1+(n-1)\times 1]\times [2+(n-1)\times 1]\times [5+(n-1)\times 1]=[1+n-1]\times [2+n-1]\times [5+n-1]=n\times (n+1)(n+4)=n^3+4n^2+n^2+4n=n^3+5n^2+4n=T_n=n^3+5n^2+4nLet{S}_n\text{be the sum to n terms of the given series; Then,}{S}_n={\sum }_{n-1}^n{T}_n={\sum }_{n-1}^n(n^3+5n^2+4n)={\sum }_{n-1}^n{n}^3+{\sum }_{n-1}^{n}5n^2+{\sum }_{n-1}^{n}4n={\sum }^n{n}^3+5{\sum }_{n-1}^{n}n={\left[\frac{n(n+1)}{2}\right]}^2+5\left[\frac{n(n+1)(2n+1)}{2}\right]+4\left[\frac{n(n+1)}{2}\right]=\frac{n^2(n+1{)}^2}{4}+\frac{5n(n+1)(2n+1)}{6}+2n(n+1)=\frac{3n^2(n+1{)}^2+10n(n+1\left)\right(2n+1)+24n(n+1)}{12}=\frac{n(n+1)}{12}[3n^2+3n+20n+10+24]=\frac{n(n+1)}{12}[3n^2+23n+34]=\frac{n(n+1)(3n^2+23n+34)}{12}Hence,S_n=\frac{n(n+1)(3n^2+23n+34)}{12}$