Question

Prove, by method of induction, for all n ∈ N:
$\rm \frac{1}{3.4.5} +\frac{2}{4.5.6}+\frac{3}{5.6.7} +...+ \frac{n}{(n+2)(n+3)(n+4)}= \frac{n(n+1)}{6(n+3)(n+4)}$

Answer 1

Solution :

Given ,

$\rm \frac{1}{3.4.5} +\frac{2}{4.5.6}+\frac{3}{5.6.7} +...+ \frac{n}{(n+2)(n+3)(n+4)}= \frac{n(n+1)}{6(n+3)(n+4)}$

Let the given statement be P(n) , i.e .,

P(n):$\rm \frac{1}{3.4.5} +\frac{2}{4.5.6}+\frac{3}{5.6.7} +...+ \frac{n}{(n+2)(n+3)(n+4)}= \frac{n(n+1)}{6(n+3)(n+4)}$

For n = 1 , we have

P(1) : 1/(3.4.5)

= [1(1+1)]/[6(1+3)(1+4)]

= 2/[6×4×5)

= 1/(3.4.5) , which is true .

Let P(k) be true for some positive

integers k , i.e.,

$\rm \frac{1}{3.4.5} +\frac{2}{4.5.6}+\frac{3}{5.6.7} +...+ \frac{k}{(k+2)(k+3)(k+4)}= \frac{k(k+1)}{6(k+3)(k+4)}$ -----( 1 )

We shall now prove that P(k+1) is true

Consider

$\rm \frac{1}{3.4.5} +\frac{2}{4.5.6}+\frac{3}{5.6.7} +...+ \frac{k}{(k+2)(k+3)(k+4)}+\frac{(k+1)}{(k+3)(k+4)(k+5)}=\frac{k(k+1)}{6(k+3)(k+4)}+\frac{(k+1)}{(k+3)(k+4)(k+5)}$

= [(k+1)]/[(k+3)(k+4)][(k/6)+1/(k+5)]

= [(k+1)]/[(k+3)(k+4)]{[k(k+5) + 6]/6(k+5)]

=[(k+1)]/[(k+3)(k+4)]{[k²+5k+6]/6(k+5)]

=[(k+1)(k+2)(k+3)]/[6(k+3)(k+4)(k+5)]

=[(k+1)(k+2)]/[6(k+4)(k+5)]

Thus , P(k+1) is true whenever P(k)

is true.

Hence , by mathematical induction,

statement P(n) is true for all natural

Numbers i.e ., n

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