Question

Prove that : $\left(1\times 2\times 5\right)+\left(2\times 3\times 7\right)+\left(3\times 4\times 9\right)+.....\left(nterms\right)$

is equal to $\frac{n\left(n+1\right)\left(n+2\right)\left(3n+7\right)}{6}$.

Answer 1

For these sort of questions i.e sequences find the general term and use the following results:

$\sum _{i=1}^{n}i=\frac{n(n+1)}{2}$

$\sum _{i=1}^n{i}^2=\frac{n(n+1)(n+2)}{6}$

$\sum _{i=1}^n{i}^3=(\frac{n(n+1)}{2}{)}^2$

general term of the given equation is:-

$r\times (r+1)\times (2r+3)=2r^3+5r^2+3r$

$\sum _{i=1}^n[r\times (r+1)\times (2r+3)]=\sum _{i=1}^n[2r^3+5r^2+3r]$

$=2[\sum _{i=1}^n{r}^3]+5[\sum _{i=1}^n{r}^2]+3[\sum _{i=1}^{n}r]$

by using above results and simplifying gives:

$=\frac{n(n+1)(n+2)(3n+7)}{6}$