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We have to prove that sum of n terms and the n+1 th term is equal to the sum according to the formula given.  This is the proof by Induction.

S_n=4\frac{1}{2*3}+8\frac{2}{3*4}+16\frac{3}{4*5}+..\ n\ terms.\\\\T_n=\frac{2^{n+1}*\ n}{(n+1)(n+2)}\\\\S_n=\frac{2^{n+2}}{n+2}-2\\\\Let\ n=1,\ \ \ 4\frac{1}{2*3}=\frac{2}{3}\\\\By\ formula\ S_1=\frac{2^3}{3}-2=\frac{2}{3,\ \ \ Formula\ holds.}\\\\Let\ S_n=\frac{2^{n+2}}{n+2}-2\ \ be\ true\ \ for\ n \ge 1\\\\S_{n+1}=\frac{2^{n+3}}{n+3}-2

S_{n+1}-S_n\\\\=\frac{2^{n+3}}{n+3}-\frac{2^{n+2}}{n+2}\\\\=\frac{(n+2)2^{n+3}-(n+3)2^{n+2}}{(n+3)(n+2)}\\\\=\frac{2^{n+2}*\ [2(n+2)-(n+3)]}{(n+3)(n+2)}\\\\=\frac{2^{n+2}*\ [n+1]}{(n+3)(n+2)}\\\\=T_{n+1}\\\\OR,\ \ S_{n+1}=S_n+T_n

Hence proved by Induction, as if the formula is good for  n, then it is good for n+1 also.

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