A problem on series . I found this problem on series in my school textbook.
If one of the two series below is 2,what is the value of the other series?
The series are:
$$(1+x+x^{2}+x^{3}+\cdots+x^{n})$$
and the other one is 
$$(x+2x^{2}+3x^{3}+\cdots+nx^{n})$$
The four options are:
A)   1
B)   2
C)   3
D)   4
I did it in the following way
We can prove
 $$ (1+x+x^{2}+x^{3}+x^{4}+\cdots+x^{n})=\frac{1}{1-x}$$
and differentiating both sides of the series we get ,
$$(x)\frac{d}{dx} (1+x+x^{2}+x^{3}+x^{4}+\cdots+x^{n})=
 (x)\frac{d}{dx}\left(\frac{1}{1-x}\right)$$
So we can say $$(x+2x^{2}+3x^{3}+4x^{4}+\cdots+nx^{n})=\frac{x}{(1-x)^{2}}$$
So assuming the second series to be equal to 2, we get,
$$\frac{x}{(1-x)^{2}}=2 $$ 
solving which we get 
$$x=\frac{5\pm\sqrt{25-16}}{4}$$
Thus the rooots of $x$ are $2 , 0.5$.
Thus putting it into the equation of the first series $$\frac{1}{1-x}$$ for x=2, we get $$\frac{1}{(1-x)}=-2$$ which is not there in the option.So IT WILL CERTAINLY NOT BE THE ANSWER .
Thus our assumption as to the second series being equals to two was wrong .
Thus now we CAN say that the first series is equal to 2.
Thus, $$\frac{1}{1-x}=2$$ and solving for $x$ we get $$x=0.5$$
Plugging in this value into the second series we get ,
$$(0.5+2(0.5)^{2}+3(0.5)^{3}+4(0.5)^{4}+\cdots+n(0.5)^{n})=\frac{0.5}{(1-0.5)^{2}} $$
We get the series being equal to $$\frac{0.5}{(0.5)^{2}}$$ which we find is INDEED EQUAL TO 2
Plz suggest any better way in which i can do the problem.
THANX WITH ALL MY HEART TO ANYONE FOR THEIR SUGGESTIONS!!

In case the sum is infinite, that is, you're considering $$1+x+x^2+\dots\tag{1}$$
Assume that $(1)$ is equal to $2$, so that $$x+x^2+x^3+\dots=1\tag{2}$$ Then look at: $$\begin{align}x+x^2+x^3+\dots\\x^2+x^3+\dots\\x^3+\dots\\\\x+2x^2+3x^3+\dots\tag{3}\end{align}$$
The desired sum $(3)$ is equal to $$x(1+x+x^2+\dots)+x^2(1+x+x^2+\dots)+\dots$$ Using $(2)$, this last sum is equal to $$2x+2x^2+2x^3+\dots=2(x+x^2+x^3+\dots)=2\cdot 1=2$$
