The given sequence is $\sum_{k=1}^{n}kx^{k}$.
How can I do this? I think I should start with the already known generating function:
$A(x) = \sum a^{n}x^{n} = \frac{1}{1 - ax}$
And then I should do something with this function so that I finally come up with the given one. But well, how can I do this?
$\sum_{k=1}^{n}kx^{k} = \sum_{k>=1} [A(x) - 1]k$
I could only get something like this, but first of all it's not the formula without sums, and second I'm not sure it's correct.
Could you help me?
$$\frac{\mathrm{d}x^{k}}{\mathrm{d}x}=kx^{k-1}$$
and$$\dfrac{d(\dfrac{1}{1-x})}{dx}=\dfrac{1}{(1-x)^{2}}$$
So when $n\rightarrow \infty$,$$1+x+x^{2}+\cdots=\dfrac{1}{1-x}$$ On differentiating, $$1+2x+3x^{2}+\cdots=\dfrac{1}{(1-x)^{2}}$$ Multiply with $x$, $$x+2x^{2}+3x^{3}+\cdots=\dfrac{x}{(1-x)^{2}}$$
An idea without (properly) generating functions:
$$\forall\,|x|<1\;,\;\;\frac1{1-x}=\sum_{k=0}^\infty x^k\stackrel{\text{derivative}}\implies\frac1{(1-x)^2}=\sum_{k=1}^\infty kx^{k-1}\;\;\ldots$$
