Finding the sum of $rx^{r-1}$

Question

May I know how to find $rx^{r-1}$. I searched for it, it said that we can utilize $\frac{d}{dx}(1-x)^{-1} =\frac{d}{dx} (1-x)^{-2}=\frac{d}{dx} \sum {x^t} =\sum {tx^{t-1}} $ Why is this true?

Thank you very much for your reply.

Answer 1

Let: $$M_n\left(x\right)=1+2x+3x^{2}+...+nx^{\left(n-1\right)}=\sum_{k=1}^{n}kx^{\left(k-1\right)}$$

then: $$\int_{ }^{ }\left(1+2x+3x^{2}+...+nx^{\left(n-1\right)}\right)dx=\int_{ }^{ }\left[\sum_{k=1}^{n}kx^{\left(k-1\right)}\right]dx=\sum_{k=1}^{n}k\int_{ }^{ }x^{\left(k-1\right)}dx=\sum_{k=1}^{n}x^{k}+C$$ on the other hand: $$\sum_{k=1}^{n}x^{k}+C=\frac{x^{n}-1}{x-1}+C$$

finally $$M_n\left(x\right)=\frac{d}{dx}\left(\frac{x^{n}-1}{x-1}+C\right)=\frac{nx^{\left(n-1\right)}\left(x-1\right)-\left(x^{n}-1\right)}{\left(x-1\right)^{2}}$$

Another way is using generating functions.

Answer 2

If you want a proof without calculus, let $$S=1+2x+3x^2+4x^3\dots+nx^{n-1}+(n+1)x^n$$ Then $$-2xS=-2x-4x^2-6x^3-\dots-2nx^n-2(n+1)x^{n+1}$$ and $$x^2S=x^2+2x^3+\dots+(n-1)x^n+nx^{n+1}+(n+1)x^{n+2}$$ Adding we get $$S(1-x)^2=1-(n+2)x^{n+1}+(n+1)x^{n+2}$$ For the infinite sum we need $|x|<1$ and the result then becomes $S=1/(1-x)^2$.