# What's the Summation formulae of the series $2*2^0 + 3*2^1 + 4*2^2 + 5*2^3…$?

I faced this question where I was asked to find a summation formulae for $n$ terms of $2*2^0 + 3*2^1 + 4*2^2 + 5*2^3.......$ I did try generalizing it with $$a_n = (n + 1)2^{n - 1}; n2^{n - 2}$$ but to no avail then I tried subtracting $2*(\sum 2^n)$ increasing geometric series from the above series( $2*2^0 + 3*2^1 + 4*2^2 + 5*2^3.......$) and I actually got something like a general term of $n2^n$ of the series obtained from this subtraction but then it lead me nowhere and I also think there is no scope for telescoping this kind of series.

I think the problem here is now I am devoid of any idea of how to approach this problem.

• Looking for the partial sum? – MCT Oct 28 '15 at 21:05
• Yeah actually I had forgotten using that term – Arnav Das Oct 28 '15 at 21:06
• Just plugging in numbers I got sum = $(n+1)*2^{n+1}$ – fleablood Oct 28 '15 at 21:20
• You can use $\sum_{n=0}^{\infty} 2^n(n+2)$ – Jan Oct 28 '15 at 21:22
• Notice $(k+2)2^k = (2k+2 - k)2^k = (k+1)2^{k+1} - k2^k$, the sum is a telescoping one. – achille hui Oct 28 '15 at 21:52

Hint. You can write the partial sums as $$s_m = \sum_{n=0}^{m} (2 + n) 2^n = \sum_{n=0}^m 2^{n+1} + \sum_{n=1}^m n 2^n$$

The first series should be easy to get a formula for using the geometric sum formula.

For the second, notice that we can write it as

\begin{align*} 2 + &4 + &8 + &16 + \dots + 2^m &= \frac{2(1 - 2^{m+1})}{1 - 2}\\ &4 + &8 + &16 + \dots + 2^m &= \frac{4(1 - 2^{m+1})}{1 - 2}\\ & &8 + &16 + \dots + 2^m &= \frac{8(1 - 2^{m+1})}{1 - 2}\\ &&&\text{etc.}\\ & & & &= \sum_{n=1}^m 2^n (2^{m+1} - 1) \end{align*}

which itself is a geometric series.

• Oh hell man I should have introspected a bit about $n*2^n$ – Arnav Das Oct 29 '15 at 4:07

Let $$S_n(x)=\sum_{k=0}^n x^k=\frac{x^{n+1}-1}{x-1}.$$

Then

$$x\frac{dS_n(x)}{dx}=\sum_{k=0}^n kx^k=x\frac{(n+1)x^n}{x-1}-x\frac{x^{n+1}-1}{(x-1)^2}.$$

With $x=2$,

$$\sum_{k=0}^n 2^k=2^{n+1}-1,\\\sum_{k=0}^n k2^k=(n+1)2^{n+1}-2^{n+2}+2,$$ and $$\sum_{k=0}^n (2+k)2^k=2\left(2^{n+1}-1\right)+(n+1)2^{n+1}-2^{n+2}+2=(n+1)2^{n+1}.$$

Check:

$$2\cdot1+3\cdot2+4\cdot4+5\cdot8+6\cdot16\to2, 8, 24,64,160\to1\cdot2,2\cdot4,3\cdot8,4\cdot16,5\cdot32$$

• You used calculus !! Well that's lovely – Arnav Das Oct 29 '15 at 4:14
• :-D ........oh for sure – Arnav Das Oct 29 '15 at 16:00

I can provide another idea: $$\begin{array}{cccccccc} a= & 2\times2^{0}+ & 3\times2^{1}+ & 4\times2^{2}+ & 5\times2^{3}+ & \cdots & \left(M+1\right)\times2^{M-1}+ & \cdots\\ 2a= & & 2\times2^{1}+ & 3\times2^{2}+ & 4\times2^{3}+ & \cdots & \left(M\right)\times2^{M-1}+ & \cdots \end{array}$$ Then, by subtracting the first equation by the second one, you can get $$-a=2\times2^{0}+2+2^{2}+2^{3}+2^{4}+\cdots+2^{M-1}+\cdots$$

If it is the finite sum, i.e., $a_M=\sum_{n=1}^{M}(n+1)2^{n-1}$, then the above equation has the last term which is negative, given by

$$-a_{M}=2\times2^{0}+2+2^{2}+2^{3}+2^{4}+\cdots+2^{M-1}-\left(M+1\right)\times2^{M}$$

I think you can get the result from here, which is $a_M = M\times2^M$.

Well, I realize this isn't clever or direct but...

$2*2^0 = 2 = 1*2^1$

$2*2^0 + 3*2^1 = 8 = 2*2^2$

$2*2^0 + 3*2^1 + 4*2^2 = 24 = 3*2^3$

So I figure $\sum_{i=0}^n(i+2)2^i = (n+1)2^{n+1}$ and I show it by induction.

$\sum_{i=0}^{n+1}(i+2)2^i= (n+1)2^{n+1} + (n+3)2^{n+1} = (2n + 4)2^{n+1} = (n+2)2^{n+2}$

Done.