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
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
He put the "k" with the x^3 part instead of with the 1/x part.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
here's a pree cool nested root integral to try out;
Woahh you do nested root integrals in general maths coughhere's a pree cool nested root integral to try out;
Was the answer to this, multiples of 4?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
YesWas the answer to this, multiples of 4?