Find the sum!

Algebra Level 3

Let S n = ∑ k = 0 n 1 k + 1 + k . S_n=\displaystyle \sum^{n}_{k=0}\dfrac{1}{\sqrt{k+1}+\sqrt{k}}. What is the value of ∑ n = 1 99 1 S n + S n − 1 ? \displaystyle \sum^{99}_{n=1}\dfrac{1}{S_n+S_{n-1}}?


The answer is 9.

View wiki
S P by s p , 16, India

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

S P
May 12, 2018

Observe that S n = ∑ k = 0 n 1 k + 1 + k = ∑ k = 0 n k + 1 − k = ( 1 − 0 ) + ( 2 − 1 ) + ( 3 − 2 ) + . . . . . . + ( n − n − 1 ) + ( n + 1 − n ) = n + 1 \begin{aligned}S_n=\displaystyle \sum^{n}_{k=0}\dfrac{1}{\sqrt{k+1}+\sqrt{k}}=\displaystyle \sum^{n}_{k=0}\sqrt{k+1}-\sqrt{k} \\& = (\sqrt{1}-\sqrt{0})+(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+......+(\sqrt{n}-\sqrt{n-1})+(\sqrt{n+1}-\sqrt{n}) \\& =\sqrt{n+1}\end{aligned}

So, S n = n + 1 S_n=\sqrt{n+1} and Similarly, S n − 1 = n S_{n-1}=\sqrt{n}

Therefore, ∑ n = 1 99 1 S n + S n − 1 = ∑ n = 1 99 1 n + 1 + n = ∑ n = 1 99 n + 1 − n = ( 2 − 1 ) + ( 3 − 2 ) + . . . . . . + ( 100 − 99 ) = 100 − 1 = 9 \begin{aligned}\displaystyle \sum^{99}_{n=1}\dfrac{1}{S_n+S_{n-1}}=\displaystyle \sum^{99}_{n=1}\dfrac{1}{\sqrt{n+1}+\sqrt{n}} \\& =\displaystyle \sum^{99}_{n=1}\sqrt{n+1}-\sqrt{n} \\& =(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+......+(\sqrt{100}-\sqrt{99}) \\& = \sqrt{100}-\sqrt{1} \\& = \boxed{9}\end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

To the last third line there should be + instead of × for the sum to Telescope.

Naren Bhandari - 3 years ago

Log in to reply

Thank you for pointing me out... Edited

s p - 3 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...