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?

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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.