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#
