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Using the principle of mathematical induction, prove each of the following for all n ϵ N:

\(\frac{1}{1.3}\) + \(\frac{1}{3.5}\) + \(\frac{1}{5.7}\) +......+ \(\frac{1}{(2n-1)(2n+1)}\) = \(\frac{n}{(2n+1)}\)

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Best answer

To Prove:

\(\frac{1}{1\times3}\) + \(\frac{1}{(3\times5)}\) +......+ \(\frac{1}{(2n-1)\times(2n+1)}\) = \(\frac{n}{(2n+1)}\)

Let us prove this question by principle of mathematical induction (PMI)

Let P(n):\(\frac{1}{1\times3}\) + \(\frac{1}{(3\times5)}\) +......+ \(\frac{1}{(2n-1)\times(2n+1)}\) = \(\frac{n}{(2n+1)}\)

For n = 1 

LHS = \(\frac{1}{1\times3}\) = \(\frac{1}{3}\)

RHS = \(\frac{1}{(2+1)}\) = \(\frac{1}{3}\)

Hence, LHS = RHS 

P(n) is true for n = 1 

Assume P(k) is true

\(\frac{1}{1\times3}\) + \(\frac{1}{(3\times5)}\) +......+ \(\frac{1}{(2k-1)\times(2k+1)}\) = \(\frac{k}{(2k+1)}\).....(1)

We will prove that P(k + 1) is true

[Writing the second last term]

(Splitting the numerator and cancelling the common factor) 

= RHS 

LHS = RHS 

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

By the principle of mathematical induction, P(n) is true for× 

Where n is a natural number 

Hence proved.

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