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Lets put 2 to be equal to x .

We define S to be x + 2 $x^{2}$ + 3 $x^{3}$ + 4 $x^{4}$ .........n $x^{n}$

Multiplying LHS and RHS by x , we obtain x S = $x^{2}$ + 2 $x^{3}$ + 3 $x^{4}$ + .....+ n $x^{n+1}$

Subtracting the second sum from the first, S(1- x ) = x + $x^{2}$ + $x^{3}$ + ....+ $x^{n}$ - (n+1) $x^{n}$

Now, apply standard GP sum formula, and after putting x = 2 equate, to obtain the answer.

We know that,
$\displaystyle \sum_{r=0}{r=n}x^{r} = \dfrac{x^{n+1} - 1}{x-1}$
Differentiating with respect to x,
$\displaystyle \sum_{r=0}^{n}rx^{r-1} = \dfrac{(x-1)nx^{n} - (x^{n+1} - 1)}{(x-1)^2}$
Multiplying by x, $\sum_{r=0}^{n}rx^{r} = x \times\dfrac{(x-1)nx^{n} - (x^{n+1} - 1)}{(x-1)^2}$
Put x = 2,
$\therefore 2^{n+10} + 2 = 2 \times \left(n\cdot2^{n} - 2^{n+1} + 1\right)$
$\therefore 2^{n+9} + 1=n\cdot2^{n} - 2^{n+1} + 1$
$\therefore 2^{n+9}=n\cdot2^{n} - 2^{n+1}$
$\therefore 2^{9} = n - 2$
$\therefore n = 2^{9} + 2$
$\therefore n = 514$

A general form

Let $s=\sum_{i=1}^{n} a^i=\frac{a^{n+1}-a}{a-1}$ $a\frac{ds}{da}=\sum_{i=1}^{n}ia^{i}=a^{n+1}(na-n-1)+a$ For our question, a=2 $\sum_{i=1}^{n}i2^{i}=2^{n+1}(n-1)+2$ $n-1=2^9$ $n=513$