Using mathematical induction prove the following: Σn k=1 k+4/k(k + 1)(k+2) = n(3n + 7)/2(n+1)(n+2)

Check for n = 1:

$\sum\limits_{k = 1}^1 {\frac{{k + 4}}{{k(k + 1)(k + 2)}} = \frac{{1 + 4}}{{1(1 + 1)(1 + 2)}} = \frac{5}{6}} = \frac{{1\left( {3 \cdot 1 + 7} \right)}}{{2(1 + 1)(1 + 2)}}$ - equality holds.

Let the equality hold for n = s, i. e

$\sum\limits_{k = 1}^s {\frac{{k + 4}}{{k(k + 1)(k + 2)}} = } \frac{{s(3s + 7)}}{{2(s + 1)(s + 2)}}$

Let us check the equality for n = s+1:

$\sum\limits_{k = 1}^{s + 1} {\frac{{k + 4}}{{k(k + 1)(k + 2)}} = } \frac{{s(3s + 7)}}{{2(s + 1)(s + 2)}} + \frac{{s + 1 + 4}}{{(s + 1)(s + 2)(s + 3)}} = \frac{{s(3s + 7)(s + 3) + 2(s + 5)}}{{2(s + 1)(s + 2)(s + 3)}} = \frac{{3{s^3} + 16{s^2} + 21s + 2s + 10}}{{2(s + 1)(s + 2)(s + 3)}} = \frac{{3{s^3} + 6{s^2} + 3s + 10{s^2} + 20s + 10}}{{2(s + 1)(s + 2)(s + 3)}} = \frac{{3s({s^2} + 2s + 1) + 10({s^2} + 2s + 1)}}{{2(s + 1)(s + 2)(s + 3)}} = \frac{{(3s + 10){{(s + 1)}^2}}}{{2(s + 1)(s + 2)(s + 3)}} = \frac{{(3s + 10)(s + 1)}}{{2(s + 2)(s + 3)}} = \frac{{(s + 1)(3(s + 1) + 7)}}{{2((s + 1) + 1)((s + 1) + 2)}}$

So, equality holds for n=s+1. Then equality holds for all $n \in N$ , Q.E.D.