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# Prove the following by using the principle of mathematical induction for all n ∈ N: (4) 1.2.3+2.3.4+...+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4

Prove the following by using the principle of mathematical induction for all $n\in N$:

Q : 4        $1.2.3+2.3.4+...+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}$

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Let the given statement be p(n) i.e.
$p(n):1.2.3+2.3.4+...+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}$
For n = 1  we have
$p(1):6=\left (\frac{1(1+1)(1+2)(1+3)}{4} \right )= \left ( \frac{1.2.3.4}{4} \right )=6$    ,   which is true

For  n = k  we have
$p(k):1.2.3+2.3.4+...+k(k+1)(k+2)=\frac{k(k+1)(k+2)(k+3)}{4} \ \ \ \ \ \ -(i)$   ,        Let's assume that this statement is true

Now,
For  n = k + 1  we have
$p(k+1):1.2.3+2.3.4+...+k(k+1)(k+2) + (k+1)(k+2)(k+3)$                                                                                                                                                                                                          $=(1.2.3+2.3.4+...+k(k+1)(k+2)) + (k+1)(k+2)(k+3)$

$=\frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3) \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3) }{4}$
$=\frac{(k+1)(k+2)(k+3)(k+4) }{4}$

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 n

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