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Prove that $x+2x^2+3x^3+4x^4+...+nx^n = \frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}$

I see that this can be written as $$\sum_{n=1}^n nx^n = n\sum_{n=1}^n x^n$$ $$\sum_{n=1}^n x^n = \frac{x(x^n-1)}{(x-1)}$$

$$\therefore n\sum_{n=1}^n x^n = n(\frac{x(x^n-1)}{(x-1)})$$

Following this I squared top and bottom so the denominator would be the same as that n the question. But I cannot seem to show the top as that in the question.

Can I have a hint on how to proceed? Also in the 1 line, is there a proof of the fact that a constant can be taken out of the summation?


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See the first answer here. – David Mitra Oct 20 '13 at 9:50

$$\sum_{n=1}^n nx^n = n\sum_{n=1}^n x^n$$ is not true, as we can not take out a dependent variable

Let $$S=x+2x^2+3x^3+4x^4+...+nx^n \ \ \ \ (1)$$

The given Series is basically Arithmetico-geometric series, see here or here

So, $$x\cdot S=x^2+2x^3+3x^4+4x^5+...+nx^{n+1}\ \ \ \ (2)$$

$$(1)-(2)\implies S(1-x)=x+x^2+\cdots+x^n-nx^{n+1}$$

Now, $$x+x^2+\cdots+x^n=x\frac{x^n-1}{x-1}$$

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why cant we take n out? Arent we allowed to take out constants from summations/integrals? Or is n not a constant in this case? – salman Oct 20 '13 at 9:57
@salman, for clarity write $$x+2x^2+3x^3+4x^4+...+nx^n=\sum_{1\le r\le n}rx^r$$ Can you take out $r$ ? – lab bhattacharjee Oct 20 '13 at 10:01
not in this scenario, since r is equal to a which is x? – salman Oct 20 '13 at 10:05
@salman, $r$ is equal to $x???$. Also, using $n$ which is also the upper limit, as the loop variable is really error-prone – lab bhattacharjee Oct 20 '13 at 10:13
why did we multiply by x at the start of the question and also why does this only work if we use $s_n = \frac{a(r^n-1)}{r-1}$ and not $s_n = \frac{a(1-r^n)}{1-r}$ – salman Oct 20 '13 at 10:28

Multiply $x + 2x^2 + 3x^3 + 4x^4+...+nx^n$ with $x-1$. You'll get:

$$(x + 2x^2 + 3x^3 + 4x^4+...+nx^n)(x-1) = x(x + 2x^2 + 3x^3 + 4x^4+...+nx^n) - (x + 2x^2 + 3x^3 + 4x^4+...+nx^n)$$

Expand it and you'll obviously end up with sort of telescopic series. Every number in the first term, except the last one will be of the form $nx^{n+1}$ and will have a "companion" of the second term of the form $(n+1)x^{n+1}$. So their difference will be $-x^{n-1}$. As we said the last of the first term and the first term of the second sequence will be "alone" and you'll end up with:

$$nx^{n+1} - x^{n-1} -...-x^2 - x$$

Now multiply again with $x-1$, so we'll have:

$$(nx^{n+1} - x^{n-1} -...-x^2 - x)(x-1) = nx^{n+1}(n-1) - (x^{n-1} +...-x^2 + x)(x-1)$$

Expand the first term. The second term is well know identity and is equal to $x^n - x$, anyway you can prove that using the same telescoping method we used previously. So we have:

$$nx^{n+2} - nx^{n-1} - x^n + x = nx^{n+2} - (n+1)x^{n+1} + x$$


Note that if the initial LHS was infinite the we could have written as:

$$\frac{1}{(x-1)^2} - 1$$

Using the formula:

$$\frac{1}{(1-x)^{k+1}} = \sum_{n=0}^{\infty} \binom{n+k}{n}x^n$$

But because it's finite series we have to add something instead of the 1 to make that boundary. Simular as we make $1 + x + x^2 + x^3 + x^4+... = \frac{1}{x-1}$ bounded with: $1 + x + x^2 + x^3+...+x^{n-1} = \frac{x^n - 1}{x-1}$

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