# Prove the following by using the principle of mathematical induction for all $n \in N$  $1.2+2.3+3.4+...+n.(n+1) \bigg[ \large\frac{n(n+1)(n+2)}{3} \bigg]$

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Let the given statement be $P(n)$, i.e.,
$P(n) : 1.2+2.3+3.4+...+n.(n+1) \bigg[ \large\frac{n(n+1)(n+2)}{3} \bigg]$
For $n = 1$, we have
$P(1) :1.2 = 2 = \large\frac{1(1+1)(1+2)}{3}$$\large\frac{1.2.3}{3}$$=3$, which is true.
Let $P(k)$ be true for some positive integer $k$, i.e.,
$= 1.2+2.3+3.4+...+k.(k+1) \bigg[ \large\frac{k(k+1)(k+2)}{3} \bigg]$----------(i)
We shall now prove that $P(k+1)$ is true.
Consider
$1.2+2.3+3.4+...+k.(k+1)+(k+1).(k+2)$
$[ 1.2+2.3+3.4+...+k.(k+1)]+(k+1).(k+2)$
$= \large\frac{k(k+1)(k+2)}{3}$$+(k+1)(k+2)\qquad [ Using (i) ] = (k+1)(k+2) \bigg( \large\frac{k}{3}$$+1\bigg)$
$= \large\frac{(k+1)(k+2)(k+3)}{3}$
$= \large\frac{(k+1)(k+1+1)(k+1+2)}{3}$
Thus, $P(k+1)$ is true whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., $n$.
answered Apr 29, 2014

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