# Thread: Sum of series problem #1

1. ## Sum of series problem #1

did the first part

how to do the next two part?

$\displaystyle {S}_{n}-\frac{1}{2}{S}_{n}= \frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+. ..+\frac{n}{2^n}$
$\displaystyle -(\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+...+\frac{n-1}{2^n}+\frac{n}{2^{n+1}})$
$\displaystyle = \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+. ..+\frac{1}{2^n}-\frac{n}{2^{n+1}}$

Where did i go wrong?

2. ## Re: Sum of series problem #1

Here's a hint, have a look at the 2nd and 3rd last terms of $\displaystyle S_n$ and $\displaystyle \frac{1}{2}S_n$

3. ## Re: Sum of series problem #1

Originally Posted by pickslides
Here's a hint, have a look at the 2nd and 3rd last terms of $\displaystyle S_n$ and $\displaystyle \frac{1}{2}S_n$
i dont understand what you mean?

$\displaystyle = \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+. ..+\frac{1}{2^n}-\frac{n}{2^{n+1}}$

where did i go wrong?

4. ## Re: Sum of series problem #1

$\displaystyle \displaystyle S_n = \frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\cdots +\frac{n-2}{2^{n-2}}+\frac{n-1}{2^{n-1}}+\frac{n}{2^n}$

Now do the same for $\displaystyle \displaystyle\frac{1}{2}S_n$

5. ## Re: Sum of series problem #1

Originally Posted by pickslides
$\displaystyle \displaystyle S_n = \frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\cdots +\frac{n-2}{2^{n-2}}+\frac{n-1}{2^{n-1}}+\frac{n}{2^n}$

Now do the same for $\displaystyle \displaystyle\frac{1}{2}S_n$
$\displaystyle =(\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+...+\frac{n-2}{2^{n-1}}+\frac{n-1}{2^n}+\frac{n}{2^{n+1}})$

6. ## Re: Sum of series problem #1

Now subtract like denominators at the end of the series.

7. ## Re: Sum of series problem #1

Originally Posted by pickslides
Now subtract like denominators at the end of the series.
$\displaystyle = \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+. ..+\frac{1}{2^n}-\frac{n}{2^{n+1}}$