Question

# lf Sn denotes the sum of first 'n' natural numbers then S1+S2x+S3x2+….…..+Snxn−1+…∞=

A
(1x)1
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B
(1x)2
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C
(1x)3
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D
(1x)4
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Solution

## The correct option is B (1−x)−3 The given question can be written as. 1+(1+2)x+(1+2+3)x2+(1+2+3+4)x3...∞ =1+3x+6x2+12x3...∞ (1−x)−n=1−n(−x)+−n(−n−1)x22!...∞ =1+nx+n(n+1)x22!...∞ Assuming x=1 Comparing the coefficients with that given in the question, we get. nx=3 ...(i) n(n+1)x22!=6 ...(ii) Substituting the value of nx in equation (i), we get 3(nx+x)=12 3+x=4 x=1 Since nx=3 n=3 If we do not substitute the value of x and only substitute the value of n in the original equation, we get 1+3x+6x2+12x3...∞=(1−x)−n =(1−x)−3

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