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 $\displaystyle \frac{1}{2}S_n$

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 $\displaystyle S_n$ 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?

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 $

Re: Sum of series problem #1

Quote:

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 $

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

Re: Sum of series problem #1

Now subtract like denominators at the end of the series.

Re: Sum of series problem #1

Quote:

Originally Posted by

**pickslides** Now subtract like denominators at the end of the series.

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