If n(2n+1)_0^1(1-x^n)^2n,dx = 1177_0^1(1-x^n)^2n+1,dx, then nN is equal to ________.

Question

If $n(2n+1)\displaystyle\int_0^1(1-x^n)^{2n}\,dx = 1177\displaystyle\int_0^1(1-x^n)^{2n+1}\,dx$, then $n\in\mathbb{N}$ is equal to \_\_\_\_\_\_\_\_.

Answer: 24

Accepted Answer

Write $(1-x^n)^{2n} = \big[1-(1-x^n)\big](1-x^n)^{2n}$... more directly, integrate $\displaystyle\int_0^1(1-x^n)^{2n+1}\,dx$ by parts (treating it as $1\cdot(1-x^n)^{2n+1}$, $u=(1-x^n)^{2n+1}$):

$$\int_0^1(1-x^n)^{2n+1}\,dx = \Big[x(1-x^n)^{2n+1}\Big]_0^1 + n(2n+1)\int_0^1x^n(1-x^n)^{2n}\,dx$$

$$= n(2n+1)\int_0^1(1-x^n)^{2n}\,dx - n(2n+1)\int_0^1(1-x^n)^{2n+1}\,dx$$

So $\big(1+n(2n+1)\big)\displaystyle\int_0^1(1-x^n)^{2n+1}dx = n(2n+1)\displaystyle\int_0^1(1-x^n)^{2n}dx = 1177\displaystyle\int_0^1(1-x^n)^{2n+1}dx$:

$$2n^2+n+1 = 1177 \implies 2n^2+n-1176=0 \implies n=24 \text{ (rejecting the negative root)}$$

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