How would you evaluate $\sum_{j=0}^t \frac{j}{2^j}$?

Question

According to WolframAlpha, this evaluates to $2^{-t} (-t + 2^{t + 1} - 2)$.

(The question isn't an exact duplicate, but the accepted answer there gives the answer to this question as well.) — Hans Lundmark, Feb 21 '17 at 07:29

score 2 · Answer 1 · answered Feb 21 '17 at 06:03

2

Thats an AGP let the sum be $S $ . Multiplying this series by $\frac {1}{2 }$ to get $\frac {S}{2} $.Now subtract the two series to get a geometric series with $r=\frac {1}{2} $ . And hope you know summation of a geometric series.

answered Feb 21 '17 at 06:03

Archis Welankar

15,910

score 2 · Answer 2 · answered Feb 21 '17 at 06:07

2

Hint:

\begin{align} g(x) = \sum\limits_{k=0} ^{t}x^k=\frac{1-x^{t+1} }{1-x}\end{align}

You can do something with $g'(x)$ to get what you need.

answered Feb 21 '17 at 06:07

Shashi

8,738

score 1 · Answer 3 · answered Feb 21 '17 at 06:12

1

The answer is basically brute force, but $\sum_{j=0}^t\frac{j}{2^j}=\sum_{j=0}^t\sum_{i=1}^j\frac{1}{2^j}=\sum_{i=1}^t\sum_{j=i}^t\frac{1}{2^j}=\sum_{i=1}^t(2^{1-i}-2^{-t})=2-2^{1-t}-t2^{-t}$, which is the same as what Wolfram Alpha told you. The last two steps are from the formula for the partial sum of a geometric series, and the reason you wanted to do this strange splitting is because you know how to calculate those.

answered Feb 21 '17 at 06:12

Pepe Silvia

1,664

Can you justify the very first equality? – Error 404 Feb 21 '17 at 06:13
Yes, notice that the term inside the sum is independent of $i$, so we're just summing $j$ terms all equal to $\frac{1}{2^j}$. – Pepe Silvia Feb 21 '17 at 06:47

How would you evaluate $\sum_{j=0}^t \frac{j}{2^j}$?

3 Answers3