pikachu975
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Where'd you get x^(4k-n)? The general term would have (2x^3)^(n-k) (1/x^k) which would be 3n - 4k wouldn't it^n \equiv \sum_{k=0}^n \binom{n}k 2^{k}(-1)^{n-k}x^{4k -n}$. Consider $x^j$ for $j\geq 0$ where $j$ is an expression of $k,n$. When $j=0$, what does this say about $k$ and $n$? $)
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- Thread starter eyeseeyou
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He put the "k" with the x^3 part instead of with the 1/x part.
(Remember, you can put it with either part.)
here's a pree cool nested root integral to try out;
pikachu975
Premium Member
Woahh you do nested root integrals in general mathshere's a pree cool nested root integral to try out;
cough
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dan964
what
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