To prove by induction we start with n=1: the first term is 1/3. The formula gives us 2/6=1/3, so the formula holds for n=1.

Assume the formula holds for the sum to the nth term. So S(n)=n(n+1)/(2(2n+1)).

Therefore S(n+1)=S(n)+(n+1)^2/((2n+2-1)(2n+2+1))=S(n)+(n+1)^2/((2n+1)(2n+3)).

Substituting for S(n): n(n+1)/(2(2n+1))+(n+1)^2/((2n+1)(2n+3))=((n+1)/(2n+1))(n/2+(n+1)/(2n+3))=

((n+1)/(2n+1))((2n^2+5n+2)/(2(2n+3))=((n+1)/(2n+1))((2n+1)(n+2)/(2(2n+3))=

(n+1)(n+2)/(2(2n+3))=S(n+1). So since the base case is satisfied (when n=1) and we can derive the (n+1)th term from the nth term, by induction the formula is correct.

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