Sum 1 + 2x + 3x & # 178;; + +Nx ^ (n-1) takes x as 2

Sum 1 + 2x + 3x & # 178;; + +Nx ^ (n-1) takes x as 2

Sn=1+2x+3x^2+-----+nX^(n-1)=
xSn= x +2x^2+...+(n-1)x^(n-1)+nx^n
Subtraction of two formulas:
(1-x)Sn=1+x+x^2+..x^(n-1)-nx^n=(1-x^n)/(1-x)-nx^n
Sn=(1-x^n)/(1-x)^2-nx^n/(1-x)

Sum Sn = x + 2x2 + 3x3 + +nxn（x≠0）．

When x = 1, Sn = 1 + 2 + 3 + +When x ≠ 0 and X ≠ 1, Sn = x + 2x2 + 3x3 + +nxn，①xSn=x2+2x3+3x4+… +N xn + 1, and (1-x) Sn = x + x2 + X3 + +Xn-nxn + 1, so Sn = x (1 − xn) (1 − x) 2-nxn + 11 − X

How to find Sn = x + 2x ^ 2 + 3x ^ 3 +. + NX ^ n (x is not equal to 0)?

x=1,Sn=1+2+…… +N = omitted
x≠1
Sn=x+2x^2+3x^3+.+nx^n
xSn=x^2+2x^3+3x^4+.+(n-1)x^n+nx^(n+1)
subtract
(x-1)Sn=-(x+x^2+x^3+.+x^n)+nx^(n-1)
=-x(x^n-1)/(x-1)+nx^(n+1)
Sn=-x(x^n-1)/(x-1)^2+nx^(n+1)/(x-1)