The sum to 50 terms of the series ( 1+2left(1+frac{1}{50}right)+3left(1+frac{1}{50}right)^{2}+4left(1+frac{1}{50}right)^{3}+ldots ldots . . ) is given by
(A) 2500
(B) 2550
(C) 2450
(D) 2400

( T_{n}=nleft(1+frac{1}{50}right)^{n-1} )
( S_{n}=angle T_{n}=sum nleft(1+frac{1}{50}right)^{n-1}=sum_{n=1}^{50} nleft(frac{51}{50}right)^{n-1}-0 )
( S_{n}left(1+frac{1}{50}right)=left(1+frac{1}{50}right) sum nleft(frac{51}{50}right)^{n-1} )
( frac{51}{50} S_{n}=sum_{n=1}^{50} nleft(frac{51}{50}right)^{n} quad-3 )
( S_{n}=1 cdotleft(frac{51}{50}right)^{0}+2left(frac{51}{50}right)+3left(frac{51}{50}right)^{2}+cdots+50left(frac{51}{50}right)^{49} )
( begin{array}{ll}frac{s 1}{50} S_{n}= & 1 cdotleft(frac{51}{50}right)+2left(frac{51}{50}right)^{2}+cdots+49left(frac{51}{50}right)^{49}+50left(frac{51}{50}right)^{50} text { Subtact both } & S_{n}-frac{51}{50} S_{n}=1+frac{51}{50}+left(frac{51}{50}right)^{2}+cdots+left(frac{5}{50}right)^{49}-50left(frac{51}{50}right)^{50}end{array} )
( -frac{S_{n}}{50}=frac{1left(left(frac{51}{50}right)^{50}-1right)-50left(frac{51}{50}right)^{50}}{frac{51}{50}-1} )
( =50left(left(frac{51}{50}right)^{30}-1right)-50left(frac{51}{50}right)^{50}=-50 )
( Rightarrow quad-frac{S_{n}}{50}=-50 quad Rightarrow quad S_{n}=50 times 50=2500 )
Option A