Question

Sum the series $\mathrm{1.2.3}+2.3.4.+3.4.5+\cdots$ to n terms.

Answer 1

We have
$T_k=\left(\right(k^{th}\text{term of}1,2,3\cdots )\times (k^{th}\text{term of 2,3,4,}\cdots )\times (k^{th}\text{term of}3,4,5,\cdots )=\{1+(k-1)\times 1\}+\{2+(k-1)\times 1\}\times \{3+(k-1)\times 1\}=k(k+1\left)\right(k+2)=(k^3+3k^2+2k).\therefore S_n={\sum }_{k=1}^n{T}_k={\sum }_{k=1}^n(k^3+3k^2+2k)={\sum }_{k=1}^n{k}^3+3{\sum }_{k=1}^n{k}^2+2\sum {k=1}^{n}k=\frac{1}{4}{n}^2(n+1{)}^2+3.\frac{1}{6}n(n+1)(2n+1)+2.\frac{1}{2}n(n+1)\{\because {\sum }_{k=1}^n{k}^3={\left\{\frac{1}{2}n\right(n+1\left)\right\}}^2,{\sum }_{k=1}^n{k}^2=\frac{1}{6}n(n+1\left)\right(2n+1),{\sum }_{k=1}^{n}k=\frac{1}{2}n(n+1\left)\right\}=\frac{1}{4}n(n+1)\left\{n\right(n+1)+2(2n+1)+4\}$
Senior Secondary School Mathematics for Class 11
$=\frac{1}{4}n(n+1)(n^2+5n+6)=\frac{1}{4}n(n+1)(n+2)(n+3).\text{Hence, the required sum is}\frac{1}{4}n(n+1)(n+2)(n+3).$