Definite Integral 2

Let's do it by Reduction Formula.

We can see it as the special case of the following integral. $I_{n} = \int_{0}^{1} (1-x^{50})^{n}dx$ We can apply by parts in it. $I_{n} = \big[x\times (1-x^{50})^{n})\big]_{0}^{1}-\int_{0}^{1}x\times n(1-x^{50})^{n-1} (-50x^{49})dx$ $I_{n} = 50n\int_{0}^{1}x^{50}(1-x^{50})^{n-1}dx$ $I_{n} = 50n\Big( \int_{0}^{1} (1-x^{50})^{n-1} dx - \int_{0}^{1}(1-x^{50})^{n}\Big)$ $I_{n} = 50nI_{n-1} - 50nI_{n}$ $(1+50n) I_{n} = 50n I_{n-1}$ $\dfrac{I_{n-1}}{I_{n}} = \dfrac{1+50n}{50n}$ Put $n = 101$ to get the required answer.

Use Beta function!