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/(1*2)+1/(2*3)+...+1/(n(n+1))=n/(n+1)#, we know for #n=1#, we have #1/(1*2)=1/2# and for #n=2#, we have #1/(1*2)+1/(2*3)=1/2+1/6=2/3=2/(2+1)#.
Hence, given statement is true for #n=1# and #n=2#.
Second Step #-# Assume it is true for #n=k#, hence
#1/(1*2)+1/(2*3)+...+1/(k(k+1))=k/(k+1)#
Now let us test it for #n=k+1# i.e.
#1/(1*2)+1/(2*3)+...+1/(k(k+1))+1/((k+1)(k+2))#
= #k/(k+1)+1/((k+1)(k+2))#
= #(k(k+2)+1)/((k+1)(k+2))#
= #(k^2+2k+1)/((k+1)(k+2))#
= #(k+1)^2/((k+1)(k+2))#
= #(k+1)/(k+2)#
Hence we see that the statement is true for #n=k+1# if it is true for #n=k#.
Hence #1/(1*2)+1/(2*3)+...+1/(n(n+1))=n/(n+1)#
