How do I prove this by mathematical induction?

1/(7*9)+1/(9*11)+1/(11*13)+... to n terms =n/(7(2n+7)

1 Answer
May 7, 2018

Please see below.

Explanation:

Induction method is used to prove a statement. Most commonly, it is used to prove a statement, involving, say n where n represents the set of all natural numbers.

Induction method involves two steps, One, that the statement is true for n=1 and say n=2. Two, we assume that it is true for n=k and prove that if it is true for n=k, then it is also true for n=k+1.

First Step - Now for 1/(7*9)+1/(9*11)+1/(11*13)+...=n/(7(2n+7)), we know for n=1, we have 1/(7*9)=1/(7(2*1+7) and for n=2, we have 1/(7*9)+1/(9*11)=2/(7(2*2+7))=2/(7*11) or (11+7)/(7*9*11)=2/(7*11)

Hence, given statement is true for n=1 and n=2.

Second Step - Here n^(th) term is 1/((2n+5)(2n+7)). Now assume it is true for n=k i.e.

1/(7*9)+1/(9*11)+1/(11*13)+...+1/((2n+5)(2n+7))=n/(7(2n+7))

Now let us test it for n=k+1 i.e..

1/(7*9)+1/(9*11)+1/(11*13)+...+1/((2n+5)(2n+7))+1/((2n+7)(2n+9))

= n/(7(2n+7))+1/((2n+7)(2n+9))

= (n(2n+9)+7)/(7(2n+7)(2n+9))

= (2n^2+9n+7)/(7(2n+7)(2n+9))

= ((2n+7)(n+1))/(7(2n+7)(2n+9))

= (n+1)/(7(2n+9)

Hence, it is true for n=k+1 and 1/(7*9)+1/(9*11)+1/(11*13)+...=n/(7(2n+7)) is true for all values of ninNN