2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Looking for the partial sum? $\endgroup$ – MCT Oct 28 '15 at 21:05
  • $\begingroup$ Yeah actually I had forgotten using that term $\endgroup$ – Arnav Das Oct 28 '15 at 21:06
  • 1
    $\begingroup$ Just plugging in numbers I got sum = $(n+1)*2^{n+1}$ $\endgroup$ – fleablood Oct 28 '15 at 21:20
  • 1
    $\begingroup$ You can use $\sum_{n=0}^{\infty} 2^n(n+2) $ $\endgroup$ – Jan Oct 28 '15 at 21:22
  • 3
    $\begingroup$ Notice $(k+2)2^k = (2k+2 - k)2^k = (k+1)2^{k+1} - k2^k$, the sum is a telescoping one. $\endgroup$ – achille hui Oct 28 '15 at 21:52
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Oh hell man I should have introspected a bit about $n*2^n$ $\endgroup$ – Arnav Das Oct 29 '15 at 4:07
1
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ You used calculus !! Well that's lovely $\endgroup$ – Arnav Das Oct 29 '15 at 4:14
  • $\begingroup$ :-D ........oh for sure $\endgroup$ – Arnav Das Oct 29 '15 at 16:00
1
$\begingroup$

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 $a_M = M\times2^M$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.