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Prove the following by using the principle of mathematical induction for all $ n \in N $ \[\] $1.2.3 + 2.3.4+...+n(n+1)(n+2)= \large\frac{n(n+1)(n+2)(n+3)}{4}$

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Let the given statement be $P(n)$, i.e.,
$P(n) : 1.2.3+2.3.4 +...+n(n+1)(n+2)= \large\frac{n(n+1)(n+2)(n+3)}{4}$
For $n=1$, we have
$P(1) : 1.2.3 = 6 = \large\frac{1(1+1)(1+2)(1+3)}{4}$$=\large\frac{1.2.3.4}{4}$$=6$ which is true.
Let $P(k)$ be true for some positive integer $k$, i.e.,
$= 1.2.3+2.3.4 +...+k(k+1)(k+2)= \large\frac{k(k+1)(k+2)(k+3)}{4}$--------(i)
We shall now prove that $P(k+1)$ is true.
Consider
$= 1.2.3+2.3.4 +...+k(k+1)(k+2)+(k+1)+(k+2)+(k+3)$
$= \{1.2.3+2.3.4 +...+k(k+1)(k+2) \}+(k+1)+(k+2)+(k+3)$
$ = \large\frac{k(k+1)(k+2)(k+3)}{4}$$+(k+1)(k+2)+(k+3) \qquad$ [ Using (i) ]
$= (k+1)(k+2)(k+3) \bigg( \large\frac{k}{4}$$+1 \bigg)$
$= \large\frac{(k+1)(k+2)(k+3) (k+4)}{4}$
$= \large\frac{(k+1)(k+1+1)(k+1+2) (k+1+3)}{4}$
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 by thanvigandhi_1
 

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