Find the sum: x + 2x^2 + 3x^3 + ... + n*x^n?

Answer 1

KNOWING THAT

1+x+x^2 + ... + x^n = (x^(n+1)-1)/(x-1)
differentiating both sides tou get

1 + 2 x + .. +n x^n-1= d/dx((x^(n+1)-1)/(x-1)

multiplying by x you get the result

LHS = x d/dx(((x^(n+1)-1)/(x-1))

Answer 2

Let S = x + 2x^2 + 3x^3 + ... + n x^n
Then x*S = x^2 + 2x^3 + ... + (n-1)x^n + n x^(n+1)

Subtract:

S - x*S = x + x² + x³ + ... + x^n - n x^(n+1)
(1-x)S = x (1 + x + x² + ... + x^(n-1)) - n x^(n+1)
(1-x)S = x [(x^n - 1)/(x-1)] - n x^(n+1)
(x-1)S = n x^(n+1) - x [(x^n - 1)/(x-1)]

S = n x^(n+1)/(x-1) - [x (x^n - 1)]/(x-1)²

provided x ≠ 1

This may be written

S = x [nx^(n+1) - (n+1)x^n + 1]/(x-1)²

Answer 3

infinity
integral test: 1/(1-1/x)^2