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Question Number 9470 by tawakalitu last updated on 09/Dec/16

prove by mathematical induction .

\frac{1}{1.3} + \frac{1}{2.5} + \frac{1}{3.7} + \dots + \frac{1}{(2 n - 1) (2 n + 1)} = \frac{n}{2 n + 1}

Commented by mrW last updated on 10/Dec/16

please check the question!

should it be following ?

\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2 n - 1) (2 n + 1)} = \frac{n}{2 n + 1}

Commented by tawakalitu last updated on 10/Dec/16

it should be like this

Answered by mrW last updated on 10/Dec/16

it is to prove

S (n) = \underset{k = 1}{\sum^{n}} \frac{1}{(2 k - 1) (2 k + 1)} = \frac{n}{2 n + 1}

for n = 1 :

S (1) = \frac{1}{1 \times 3} = \frac{1}{3} \overset{!}{=} \frac{1}{2 \times 1 + 1} = \frac{1}{3}

\Rightarrow {it}^{'} s true

suppose {it}^{'} s true for n, we have for n + 1

S (n + 1) = S (n) + \frac{1}{[2 (n + 1) - 1] [2 (n + 1) + 1]}

= S (n) + \frac{1}{(2 n + 1) [2 (n + 1) + 1]}

= \frac{n}{2 n + 1} + \frac{1}{(2 n + 1) [2 (n + 1) + 1]}

= \frac{n [2 (n + 1) + 1] + 1}{(2 n + 1) [2 (n + 1) + 1]}

= \frac{{2 n}^{2} + 3 n + 1}{(2 n + 1) [2 (n + 1) + 1]}

= \frac{(2 n + 1) (n + 1)}{(2 n + 1) [2 (n + 1) + 1]}

= \frac{(n + 1)}{2 (n + 1) + 1}

that means {it}^{'} s also true for n + 1 .

\Rightarrow {it}^{'} s true for all n ⩾ 1 .

Commented by tawakalitu last updated on 10/Dec/16

God bless you sir . i really appreciate your

effort .