First four terms of $ \sum_{k=1}^\infty \frac{(-1)^{k+1(2^k)}}{(2k - 1)!} $
By Mia Morrison •
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I'm taking single variable calculus on Coursera platform and asked :
Write first four terms of the sum :
$$ \sum_{k=1}^\infty \frac{(-1)^{k+1(2^k)}}{(2k - 1)!} $$
I'm including a screenshot of the question case my MathJax is incorrect (particularly the formatting of $2^k$) :
Here is my calculation of term1 :
Term1, taking k = 1 : $$ \frac{(-1)^{1+1(2^1)}}{(2(1) - 1)!} $$
=
$$ \frac{(-1)^{3}}{1} $$
=
-1
Is there an oversight here on my part as I know calculating term1 = -1 and as just one option (4th option) has -1 as first term in order to select correct answer can just select this option, omitting the need to calculate remaining three terms ?
$\endgroup$ 11 Answer
$\begingroup$It is $$\dfrac{(-1)^{k+1}2^k}{2k-1}$$ and not $$\dfrac{(-1)^{(k+1)2^k}}{2k-1}$$
$\endgroup$More in general
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