[sequence summation] 1 + 2x + 3x ^ 2 + +nx^n-1=()? 1+2*x+3*(x^2)+…… +n*x^(n-1)=( When x is not equal to 1, = (1-x ^ n) / [(1-x) ^ 2] - [n * (x ^ n)] / (1-x) When x equals 1, = n (n + x) / 2

[sequence summation] 1 + 2x + 3x ^ 2 + +nx^n-1=()? 1+2*x+3*(x^2)+…… +n*x^(n-1)=( When x is not equal to 1, = (1-x ^ n) / [(1-x) ^ 2] - [n * (x ^ n)] / (1-x) When x equals 1, = n (n + x) / 2

The sum of an equation is required
When it is found that the first half of the formula is constant with equal difference and the second half is equal ratio
Generally, we use dislocation subtraction
That is to multiply every term of this formula by X
Write it at the bottom of this equation
And then subtract the one above from the one below
Just simplify
This is very annoying, but it needs patience. Try more