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So I have got this practice problem

$$\sum_{k=1}^n \frac{(-1)^k\cdot2^{2k}}{3}$$

Now I am fairly new to the sigma notation for the sum of elements. I know the basic stuff like the sum of the arithmetic and geometric sequence. However with this example, I'm not sure what am I supposed to do with it. I can't even determine whether this is an arithmethic or geometric sequence.

Now of course with this little number of element, I could just sum them up manually but I'm sure that there is much easier way. It's just that I don't see it.

Would you be so king give me some hints on what am I looking at?

Thank you for any input.

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    $\begingroup$ This is a geometric series with quotient $-4$ (divided by $3$). Just apply the formula. $\endgroup$
    – Peter
    Sep 13, 2021 at 13:45
  • $\begingroup$ Note however that the series does not start with $1$ which makes it slightly more complicated. $\endgroup$
    – Peter
    Sep 13, 2021 at 13:48

1 Answer 1

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You have a geometric series with common ratio $r=-4$ and scale factor $a=\frac{1}{3}$. See here also. Then the sum of the first $n$ terms is given by

$$\sum_{k=1}^{n}ar^{k-1}=\frac{a(1-r^{n})}{1-r} \quad (1)$$

for $r\neq 1$. Applying this to the above series we obtain

$$\sum_{k=1}^{n}\frac{(-1)^{k}2^{2k}}{3}=\sum_{k=1}^{n}\frac{(-1)^{k}4^{k}}{3}=\sum_{k=1}^{n}\frac{(-4)^{k}}{3}$$

Note that the general term in $(1)$ is $ar^{k-1}$ and at $k=1$ it gives $a$, but in our case, $k=1$ gives $-\frac{4}{3}$. So we can pull out a factor of $-4$ and we have

$$\sum_{k=1}^{n}\frac{(-1)^{k}2^{2k}}{3}=-4\sum_{k=1}^{n}\frac{1}{3}(-4)^{k-1}=-4\frac{\frac{1}{3}(1-(-4)^{n})}{1-(-4)}=\frac{4}{15}((-4)^{n}-1)$$


Where does the above formula come from?

Consider again your example. Let

$$S=\sum_{k=1}^{n}\frac{(-1)^{k}2^{2k}}{3}=-\frac{2^{2}}{3}+\frac{2^4}{3}-\frac{2^6}{3}+...+\frac{(-1)^{n}2^{2n}}{3}$$ $$=\frac{1}{3}\left(-2^2+2^4-2^6+...+(-1)^n2^{2n}\right)=\frac{1}{3}\left(-4+4^{2}-4^{3}+...+(-1)^{n}4^{n}\right)$$ $$=\frac{4}{3}\left(-1+4-4^{2}+...+(-1)^{n}4^{n-1}\right)$$

Then multiply $S$ by $-4$ to obtain

$$-4S=\frac{4}{3}\left(4-4^{2}+4^{3}+...+(-1)^{n+1}4^{n}\right)$$

Next subtract $-4S$ from $S$ to obtain

$$S-(-4S)=\frac{4}{3}\left(-1+4-4^{2}+...+(-1)^{n}4^{n-1}\right)-\frac{4}{3}\left(4-4^{2}+4^{3}+...+(-1)^{n+1}4^{n}\right)$$ $$=\frac{4}{3}\left(-1+\color{red}{4}-\color{green}{4^{2}}+...+\color{blue}{(-1)^n4^{n-1}}-\color{red}{4}+\color{green}{4^2}+...-\color{blue}{(-1)^{n}4^{n-1}}+(-1)^{n}4^{n}\right)$$ $$=\frac{4}{3}\left((-1)^{n}4^n-1\right)$$ $$5S=\frac{4}{3}\left((-4)^n-1\right)\implies S=\frac{4}{15}\left((-4)^n-1\right)$$

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    $\begingroup$ @Jasasul I think the most important part of this answer is the second half, where you replace the $\Sigma$ by a sequence of summands with an ellipsis (the ...). Then you can actually see what's going on. That's the way to start this kind of question. $\endgroup$ Sep 13, 2021 at 14:58
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    $\begingroup$ @Ethan Bolker Thanks for the comment. Agreed, I just showed how to use the general formula in the first half. $\endgroup$
    – Alessio K
    Sep 13, 2021 at 15:05
  • $\begingroup$ Woah. Thank you for this in depth explanation. Really appreciate it. $\endgroup$
    – Jasasul
    Sep 13, 2021 at 19:15
  • $\begingroup$ @Jasasul You're very welcome! :-) $\endgroup$
    – Alessio K
    Sep 13, 2021 at 19:16

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