how can I calculate the sum of the n first elements : $ 1\cdot 2 +2\cdot2^2 +3\cdot2^3+4\cdot2^4+\cdots+k\cdot2^k+\cdots?$

Question

how can I calculate the sum of the n first elements :

$$ 1\cdot 2 +2\cdot2^2 +3\cdot2^3+4\cdot2^4+\cdots+k\cdot2^k+\cdots$$

I don't have idea although I tried some tricks.

I will be happy for help from you

Answer 1

Hint:

• Write the polynomial $p(x)=1+x+x^2+\cdots+x^n=\frac{x^{n+1}-1}{x-1}$
• Take the derivative $p'(x)=1+2x+3x^2+\cdots+nx^{n-1}=\ldots\text{ (can you work this one out?)}$
• Substitute $x=2$ and multiply by $2$ to get the exact expression you are looking for.

Answer 2

Simple answers are already provided which are most general. There is another approach:

$$S = \sum_{k=1}^{n} k \cdot 2^k\\ 2S = \sum_{k=1}^{n} k\cdot 2^{k+1} = \sum_{k=1}^{n} (k+1)\cdot 2^{k+1} - \sum_{k=1}^{n}2^{k+1} = S-2+(n+1)\cdot 2^{n+1} - \sum_{k=1}^{n}2^{k+1}$$

Now use the formula of Gp to get $\sum_{k=1}^{n}2^{k+1} = 4(2^{n}-1)$

$$S = -2 + n\cdot 2^{n+1}+2^{n+1}-2^{n+2}+4 \\ = 2 + (n-1)\cdot2^{n+1}$$

Answer 3

$$\sum_{k=1}^nk2^k=2\sum_{k=1}^nk2^{k-1}=2\left(\sum_{k=1}^nx^n\right)'_{x=2}=2\left(\frac{x^{n+1}-1}{x-1}\right)'_{x=2}=$$ $$=2((n+1)2^n-2^{n+1}+1)=n2^{n+1}-2^{n+1}+2.$$