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Question Number 140890 by mathsuji last updated on 13/May/21

1+2x+3x^2 +4x^3 +...+(n+1)x^n =?

1+2x+3x2+4x3+...+(n+1)xn=?

Answered by ajfour last updated on 14/May/21

x+x^2 +x^3 +...+x^(n+1) =((1−x^(n+2) )/(1−x)) differentiating wrt x 1+2x+3x^2 +...+(n+1)x^(n+1) =(((x−1)(n+2)x^(n+1) +1−x^(n+2) )/((1−x)^2 )) =((1−(n+2)x^(n+1) +(n+1)x^(n+2) )/((1−x)^2 ))

x+x2+x3+...+xn+1=1−xn+21−xdifferentiatingwrtx1+2x+3x2+...+(n+1)xn+1=(x−1)(n+2)xn+1+1−xn+2(1−x)2=1−(n+2)xn+1+(n+1)xn+2(1−x)2

Commented by mathsuji last updated on 15/May/21

thaks Sir

thaksSir

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