Find out the answer to "So I have got this practice problem∑k=1n(−1)k⋅22k3"

Cormac Gilbert

Answered question

2022-02-25

So I have got this practice problem

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

blokova5u8

Beginner2022-02-26Added 7 answers

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} a r^{k - 1} = \frac{a (1 - r^{n})}{1 - r}

(1)
for

r \neq 1

. Applying this to the above series we obtain

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

a r^{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^{2 k}}{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)

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