If First term of G.P is 1 and common ratio '1/2' then find the infinite sum of G.P

Explanation

Let `P(n):(1^2)/(1:3)+(2^2)/(3.5)+.....+(n^2)/((2n-1)(2n+1))=(n(n+1))/(2(2n+1))` .....(i)
Step I For n=1.
LHS of Eq. (i) `(1^2)/(1.3)=(1)/(3)`
RHS of Eq. (i) `=(1(1+1))/(2(2xx1+1))=(2)/(2(3))=(1)/(3)`
LHS=RHS
Therefore , P(1) is true.
Step II Let us assume that the result is true for `n=k`, then
`P(k)=(1^2)/(1.3)+(2^2)/(3.5)+....+(k^2)/((2k-1)(2k+1))=(k(k+1))/(2(2k+1))`
Step III For `n=k+1`, we have to prove that
`P(k+1):(1^2)/(1:3)+(2^2)/(3.5)+......+(k^2)/((2k-1)(2k+1))+((k+1)^2)/((2k+1)(2k+3))=((k+1)(k+2))/(2(2k+3))`
LHS `=(1^2)/(1.3)+(2^2)/(3.5)+......+(k^2)/((2k-1)(2k+1))+((2k+1)^2)/((2k+1)(2k+1))`
`=(k(k+1))/(2(2k+1))+((k+1)^2)/((2k+1)(2k+3))` [ by assumption step]
`=((k+1))/((2k+1)){(k)/(2)+(k+1)/((2+3))}=((k+1))/((2k+1)){(2k^2+5k+2)/(2(2k+3))}`
`=((k+1))/((2k+1)).((k+2)(2k+1))/(2(2k+3))=((k+1)(k+2))/(2(2k+3))=RHS`
This shows that ,the result is true for `n=k+1`. Therefore , by the principle of mathematical induction the result is true for all ` n in N`.