How can i calculate this serias?
$$\sum_{i=1}^n \frac{2^m}{2^i}\cdot i$$
I tried to do:
$$\ 2^m\cdot\left[\frac{1}{2}+\frac{2}{2^2}+...+\frac{n}{2^n}\right]$$
And i don't know how to continue..
Thank you.
Asked
Active
Viewed 88 times
4
-
Take a look here to know methods to solve this summation easily. – Masacroso Mar 27 '17 at 13:48
2 Answers
0
$$ \frac{d}{da}a^{-i} = -ia^{-i-1}\implies -a\frac{d}{da}a^{-i} = ia^{-i} $$ we can write your sum as $$ a^m\sum_{i=1}^n -a\frac{d}{da}a^{-i} = -a^{m+1}\frac{d}{da}\sum_{i=1}^n \left(a^{-1}\right)^i $$ Can you take it from here and replace $a$ with $2$?
Chinny84
- 14,186
- 2
- 22
- 31
0
$$s(n)=\frac{1}{2}+\frac{2}{2^2}+...+\frac{n}{2^n}\quad (1)$$
multiply by $1/2$ and get
$$\frac{s(n)}{2}=\frac{1}{2^2}+\frac{2}{2^3}+...+\frac{n-1}{2^{n}}+\frac{n}{2^{n+1}}\quad (2)$$
Now make $(1)-(2)$ and get
$$\frac{s(n)}{2}=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n}}-\frac{n}{2^{n+1}}\\ s(n)=\left[\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{n-1}}\right]-\frac{n}{2^{n}}$$
Can you finish?
Arnaldo
- 21,342