# Sum of series problem #1

• December 5th 2011, 11:49 AM
BabyMilo
Sum of series problem #1
did the first part

how to do the next two part?

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

Where did i go wrong?
• December 5th 2011, 11:58 AM
pickslides
Re: Sum of series problem #1
Here's a hint, have a look at the 2nd and 3rd last terms of $S_n$ and $\frac{1}{2}S_n$
• December 5th 2011, 12:04 PM
BabyMilo
Re: Sum of series problem #1
Quote:

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

i dont understand what you mean?

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

where did i go wrong?
• December 5th 2011, 12:09 PM
pickslides
Re: Sum of series problem #1
$\displaystyle S_n = \frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\cdots +\frac{n-2}{2^{n-2}}+\frac{n-1}{2^{n-1}}+\frac{n}{2^n}$

Now do the same for $\displaystyle\frac{1}{2}S_n$
• December 5th 2011, 12:11 PM
BabyMilo
Re: Sum of series problem #1
Quote:

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

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

$=(\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+...+\frac{n-2}{2^{n-1}}+\frac{n-1}{2^n}+\frac{n}{2^{n+1}})$
• December 5th 2011, 12:15 PM
pickslides
Re: Sum of series problem #1
Now subtract like denominators at the end of the series.
• December 5th 2011, 12:17 PM
BabyMilo
Re: Sum of series problem #1
Quote:

Originally Posted by pickslides
Now subtract like denominators at the end of the series.

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