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