Question

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

Answer 1

Given

$(1\times 2\times 5)+(2\times 5\times 7)+(3\times 4\times 9)+......\infty =\frac{n(n+1)(n+2)(3n+7)}{6}$

take $n=1$

LHS:

$=1\times 2\times 5$

$=10$

RHS:

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

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

$=10$

LHS=RHS

take $n=k$

LHS:

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

RHS:

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

LHS=RHS

Now to prove that $n=k+1$ is also true

LHS:

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

RHS:

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

$=\frac{(k+1)(k+2)\left(\right(k+3\left)\right(3k+10)}{6}$

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

Therefore $(1\times 2\times 5)+(2\times 5\times 7)+(3\times 4\times 9)+......\infty =\frac{n(n+1)(n+2)(3n+7)}{6}$

Hence proved by induction