Question

Help !

Answer 1

It is known that for \(\large x \neq 1\)
\(\LARGE 1+x+x^2+...x^{n-1}=\frac{1-x^n}{1-x}\),hence find the sum of the series
\[\Huge 1+2x+3x^2+..+(n-1)x^{n-2}\]

Answer 2

its easy bro.. just take derivative of both sides of given equation so we taking derivative of LHS of given equation with respect to x we get 1+2x+3x..... (n-1)x^(n-2) which must be equal to derivative of 1-x^n/1-x with respect to x
Use for quotient Rule for derivation of 1-x^n/1-x

Answer 3

it was 1+2x+3x^2...... hope you got the concept

Answer 4

\[\large 1+2x+3x^2+...+nx^{n-2}=\frac{(1-x)(-nx^{n-1})-(1-x^n)(-1)}{(1-x)^2}\]
\[\LARGE =\frac{-(1-x)(nx^{n-1})+(1-x^n)}{(1-x)^2}\]

Answer 5

That's right.. bro.. :) and sorry for poor English..