Question

Prove that $\left(1\times 2\times 5\right)+\left(2\times 3\times 7\right)+\left(3\times 4\times 9\right)+.....$ is $\frac{n\left(n+1\right)\left(n+2\right)\left(3n+7\right)}{6}$

Answer 1

$(1\times 2\times 5)+(2\times 3\times 7)+(3\times 4\times 9)+\cdots$

$p\left(n\right)=(1\times 2\times 5)+(2\times 3\times 7)+\cdot +[n\times (n+1)\times (2n+3\left)\right]$

$n=1$

$Lhs=1\times \left(2\right)\times \left(5\right)=10$

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

$=\frac{2\times 3\times 10}{6}=10$

$\text{true for}n=1$

$\text{then true for}k,$

$sop\left(k\right)=(1\times 2\times 5)+\cdot +k(k+1)(2k+3)$

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

$Nowp(k+1)=(3k^4+16k^3+27k^2+14k)/6$

$=\left[\right(1\times 2\times 5)+(2\times 3\times 7)+\cdots +k(k+1\left)\right(2k+3\left)\right]$

$+(k+1)(k+2)(2k+5)$

$=\frac{k(k+1)(k+2)(3k+7)}{6}+(k+1)(k+2)(2k+5)$

$=\frac{3k^4+16k^3+27k^2+14k+12k^3+66k^2+114k+60}{6}$

$=\frac{3k^4+28k^3+93k^2+128k+60}{6}$

$\text{hence true for}k+1$

$\therefore p\left(n\right)\text{is true}\forall n\in IN$