3
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Show that

$1,3,6,10,15,...$ for $n=1,2,3,...$ its n-th Triangular number

$$\int_{0}^{1}(1-\sqrt{x})^nd ={1\over T_n}$$

$$\int_{0}^{1}(1-\sqrt{1-x})^nd ={1\over T_n}$$

My try: using binomial theorem

$$(1-\sqrt{x})^n=1-nx^{1\over2}+{n(n+1)\over2!}x-{n(n+1)(n+2)\over3!}x^{3\over 2}-\cdots$$

$${1\over T_n}=\int_{0}^{1}\left(1-nx^{1\over2}+{n(n+1)\over2!}x-{n(n+1)(n+2)\over3!}x^{3\over 2}-\cdots\right)dx$$

$${1\over T_n}=x-n{2\over3}x^{3/2}+{n(n+1)\over2!}{x^2\over2}-{n(n1)(n+2)\over3!}{2x^{5/2}\over5}+\cdots|_{0}^{1}$$

$${1\over T_n}=1-n{2\over3}+{n(n+1)\over2!}{1\over2}-{n(n+1)(n+2)\over3!}{2\over5}+\cdots$$

This look very long and tedious.

How can I show (1) and (2) are equal using a quick technique?

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  • $\begingroup$ The substitution $u = 1 - x$ is jumping out of the page. Have you tried it yet? $\endgroup$ – Kaynex Dec 28 '16 at 17:06
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By the change of variable $x \to 1-x$ one gets the first identity.

Then by the change of variable $x=u^2$, $dx=2u\,du$, then $v=1-u$, one has $$ \int_{0}^{1}(1-\sqrt{x})^ndx=2 \int_{0}^{1}u(1-u)^ndu=2\int_{0}^{1}(1-v)v^ndu=\frac{2}{(n+1)(n+2)}=\frac1{T_n} $$ as announced.

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