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How to solve the following sum?

$$\sum_{n=0}^k (-1)^n (1/2)^n$$

andrepcg
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    It's a geometric series. Do the usual thing. – Chappers Jun 21 '15 at 13:38
  • What about $\sum_{n=0}^k (-1)^n x^n=\sum_{n=0}^k (-x)^n$ first ? – Claude Leibovici Jun 21 '15 at 13:38
  • That's a rather classical problem. What have you tried? –  Jun 21 '15 at 13:38
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    We have $(-1)^n (1/2)^n = (-1/2)^n$. – Arthur Jun 21 '15 at 13:38
  • Hint: What is the $n$-th term for a geometric sequence. – Aleksandar Jun 21 '15 at 13:40
  • @Aleksandar Sorry but I fail to get your hint. – Did Jun 21 '15 at 14:06
  • See http://math.stackexchange.com/questions/11703/proof-of-the-formula-1xx2x3-cdots-xn-fracxn1-1x-1 and http://math.stackexchange.com/questions/29023/value-of-sum-limits-n-xn – Martin Sleziak Jun 22 '15 at 05:05
  • Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote to reopen this. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – Martin Sleziak Jun 22 '15 at 05:07

3 Answers3

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HINT:

Using Exponent Combination Law $a^mb^m=(ab)^m$ for real $a>0,b>0,m$

$\displaystyle\left(\dfrac12\right)^n=\left(-\dfrac12\right)^n$ $\displaystyle\implies\sum_{n=0}^k (-1)^n \left(\dfrac12\right)^n =\sum_{n=0}^k\left(-\dfrac12\right)^n$

which is a Geometric Series with common ratio $=-\dfrac12$

lab bhattacharjee
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use the geometric serie for $|x|<1$ $$\sum_{n=0}^{k }x^n=1+x+x^2+x^3+....=\frac{1-x^{k+1}}{1-x}$$

then use $x=-1/2$

0

We have,

$s_{n}=\sum_{k=0} ^n (-\frac{1}{2})^n$

$s_0=1$

$s_1=\frac{1}{2}$

$s_2=\frac{3}{4}$

$s_3=\frac{5}{8}$

$s_4=\frac{11}{16}$

$s_5=\frac{21}{32}$

$s_{6}=\frac{43}{64}$

We can see the numerator is part of the Jacobsthal sequence.

We can later conclude,

$s_{n}=\frac{1}{3} [(-\frac{1}{2})^n +2]$

Hope this helps.

Aleksandar
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