Question

Given that; 1.3+2.4+....+n(n+2)={n(n+1)(2n+7)}÷6 Prove by induction for all integral values of n

Answer 1

Claim: $1\times 3 + 2 \times 4 + \dots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$ for all $n \in \mathbb{N}$.

Proof:

We proceed by induction. First we verify the claim for the $n=1$ case:

$1\times 3 = \frac{1(1+1)(2+7)}{6} = \frac{2\times9}{6} = \frac{18}{6} = 3.$

Clearly the $n=1$ case holds.

Now assume the claim is true for the $n=k$ case and that the equation

$1\times 3 + 2 \times 4 + \dots + k(k+2) = \frac{k(k+1)(2k+7)}{6}$

holds true.

Now we proceed to the induction step and try to show that the $n=k$ being true implies the $n=k+1$ case.

We have that

$1\times 3 + 2 \times 4 + \dots + k(k+2) + (k+1)(\{k+1\}+2) = \frac{k(k+1)(2k+7)}{6} + (k+1)(k+3)$

by the induction step.

Summing the expression on the right hand side of the equation yields

$\frac{k(k+1)(2k+7)+6(k+1)(k+3)}{6} = \frac{(k+1)(2k^2+7k)+(k+1)(6k+18)}{6} = \frac{k(k+1)(2k+7)+(k+1)(6k+18)}{6}$

which is equal to

$\frac{(2k^2+13k+18)(k+1)}{6}$

factorizing the quadratic factor ($2k^2+13k+18$) gives

$\frac{(k+2)(2k+9)(k+1)}{6} = \frac{(k+1)(\{k+1\}+1)(2\{k+1\}+7)}{6}$

Hence the equation holds for $n=k+1$ whenever it holds for $n=k$. As it holds for $n=1$ then it also holds for all $n \in \mathbb{N}$ by induction. $\blacksquare$