What could it be?

Level 2

1 1 + 2 + 1 2 + 3 + . . . + 1 2015 + 2016 = a b \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2015}+\sqrt{2016}}=\sqrt{a}-b If a a is square-free find a + b a+b

2018 2015 2016 2017

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Tom Engelsman
May 9, 2021

The series can be written as:

Σ n = 1 2015 1 n + n + 1 \large \Sigma_{n=1}^{2015} \frac{1}{\sqrt{n}+\sqrt{n+1}} ;

or Σ n = 1 2015 1 n + n + 1 n + 1 n n + 1 n ; \large \Sigma_{n=1}^{2015} \frac{1}{\sqrt{n}+\sqrt{n+1}} \cdot \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}};

or Σ n = 1 2015 n + 1 n ( n + 1 ) n ; \large \Sigma_{n=1}^{2015} \frac{\sqrt{n+1}-\sqrt{n}}{(n+1)-n};

or Σ n = 1 2015 n + 1 n \large \Sigma_{n=1}^{2015} \sqrt{n+1}-\sqrt{n}

which telescopes to 2016 1 2016 + 1 = 2017 . \sqrt{2016} - 1 \Rightarrow 2016+1=\boxed{2017}.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...