P(n)=1+4+7+.......+n(n+1)(n+2)= {n(n+1)(n+2)(n+3)}÷4 Given for P(1) and …

Let P(n) = {tex}1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + .... + n(n + 1)(n + 2){/tex}{tex} = \frac{{n(n + 1)(n + 2)(n + 3)}}{4}{/tex}
For n = 1
{tex}P(1) = 1 \times 2 \times 3 = \frac{{1 \times 2 \times 3 \times 4}}{4} \Rightarrow 6 = 6{/tex}
{tex}\therefore {/tex} P (1) is true
Let P(n) be true for n = k.
{tex}\therefore P(k) = 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ...{/tex}{tex} + (k + 1)(k + 2) = \frac{{k(k + 1)(k + 2)(k + 3)}}{4}{/tex} ... (i)
For n = k + 1
{tex}P(k + 1) = 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ... + {/tex}{tex}k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3){/tex}
{tex} = \frac{{k(k + 1)(k + 2)(k + 3)}}{4}{/tex} + (k + 1) (k + 2) (k + 3) [Using (i)]
{tex} = (k + 1)(k + 2)(k + 3)\left[ {\frac{{k + 4}}{4}} \right]{/tex}
{tex} = \frac{{(k + 1)(k + 2)(k + 3)(k + 4)}}{4}{/tex}
{tex}\therefore {/tex} P(k + 1) is true.
Thus P(k) is true {tex} \Rightarrow {/tex} P (k + 1) is true.
Hence by principle of mathematical induction, P(n) is true for all {tex}n \in N{/tex}.