Summation

Algebra Level 3

I = k = 1 2016 1 k + 1 + k I=\sum_{k=1}^{2016}{\dfrac{1}{\sqrt{k+1}+\sqrt{k}}}

Find I \lceil I \rceil .

Notation : \lceil \cdot \rceil denotes the ceiling function .


The answer is 44.

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.

2 solutions

Armain Labeeb
Jul 12, 2016

Relevant wiki: Telescoping Series - Sum

I = k = 1 2016 1 k + 1 + k Question = k = 1 2016 1 k + 1 + k ( k + 1 k k + 1 k ) Rationalize the denominator = k = 1 2016 1 ( k + 1 k ) ( k + 1 + k ) ( k + 1 k ) Multiply fractions = k = 1 2016 k + 1 k ( k + 1 + k ) ( k + 1 k ) 1 × a = a = k = 1 2016 k + 1 k ( k + 1 ) 2 ( k ) 2 ( a + b ) ( a b ) = a 2 b 2 = k = 1 2016 k + 1 k k + 1 k Simplify and cancel = k = 1 2016 k + 1 k Simplify = ( 2017 2016 ) + ( 2016 2015 ) + + ( 3 2 ) + ( 2 1 ) = 2017 2016 + 2016 2015 + + 3 2 + 2 1 Remove brackets and cancel = 2017 1 Simplify 43.9 I = 43.9 Use ceiling function = 44 \large \begin{aligned} I & =\, \, \sum _{ k=1 }^{ 2016 }{ \frac { 1 }{ \sqrt { k+1 } +\sqrt { k } } } & \text{Question} \\ & =\, \, \sum _{ k=1 }^{ 2016 }{ \frac { 1 }{ \sqrt { k+1 } +\sqrt { k } } \left( \frac { \sqrt { k+1 } -\sqrt { k } }{ \sqrt { k+1 } -\sqrt { k } } \right) } & \text{Rationalize the denominator} \\ & =\, \, \sum _{ k=1 }^{ 2016 }{ \frac { 1\left( \sqrt { k+1 } -\sqrt { k } \right) }{ \left( \sqrt { k+1 } +\sqrt { k } \right) \left( \sqrt { k+1 } -\sqrt { k } \right) } } & \text{Multiply fractions} \\ & =\, \, \sum _{ k=1 }^{ 2016 }{ \frac { \sqrt { k+1 } -\sqrt { k } }{ \left( \sqrt { k+1 } +\sqrt { k } \right) \left( \sqrt { k+1 } -\sqrt { k } \right) } } & \color{#BBBBBB}{1 \times a = a} \\ & =\, \, \sum _{ k=1 }^{ 2016 }{ \frac { \sqrt { k+1 } -\sqrt { k } }{ \left( \sqrt { k+1 } \right) ^{ 2 }-\left( \sqrt { k } \right) ^{ 2 } } } & \color{#BBBBBB}{(a+b)(a-b)=a^{2}-b^{2}} \\ & =\, \, \sum _{ k=1 }^{ 2016 }{ \frac { \sqrt { k+1 } -\sqrt { k } }{ { k }+1-k } } & \text{Simplify and cancel} \\ & =\, \, \sum _{ k=1 }^{ 2016 }{ { \sqrt { k+1 } -\sqrt { k } } } & \text{Simplify} \\ & =\, \, \, \, \left( \sqrt { 2017 } -\sqrt { 2016 } \right) +\left( \sqrt { 2016 } -\sqrt { 2015 } \right) +\dots +\left( \sqrt { 3 } -\sqrt { 2 } \right) +\left( \sqrt { 2 } -\sqrt { 1 } \right) & \\ & =\, \, \, \, \sqrt { 2017 } -\sqrt { 2016 } + \sqrt { 2016 } -\sqrt { 2015 } +\dots + \sqrt { 3 } -\sqrt { 2 } + \sqrt { 2 } -\sqrt { 1 } & \text{Remove brackets and cancel} \\ & =\, \, \, \, \sqrt { 2017 } -1 & \text{Simplify} \\ & \approx \, \, \, 43.9 & \\ \left\lceil I \right\rceil \, \, & =\, \, \, \, \, \left\lceil 43.9 \right\rceil & \text{Use ceiling function} \\ & =\, \, \, \, \, \boxed { 44 } & \end{aligned}

Nice and detailed. You've improved on LaTeX writing. Keep up the effort!

Hung Woei Neoh - 4 years, 11 months ago

Log in to reply

StackExchange and Wikipedia helped. Also, I use a Google Chrome extension called Daum Equation Editor. I like your colourful solutions though, very innovative.

Armain Labeeb - 4 years, 11 months ago

Log in to reply

Thanks. Colorful solutions are fun to write!

Hung Woei Neoh - 4 years, 11 months ago

celling function . i do the same

Abdullah Ahmed - 4 years, 11 months ago

Nicely structured!

Ashish Menon - 4 years, 11 months ago
Shikhar Mohan
Jul 11, 2016

Relevant wiki: Telescoping Series - Sum

This is a rather simple one. 1 k + 1 + k = k + 1 k ( k + 1 ) 2 ( k ) 2 = k + 1 k \frac{1}{\sqrt{k+1}+\sqrt{k}}=\frac{\sqrt{k+1}-\sqrt{k}}{({\sqrt{k+1})}^{2}-({\sqrt{k}})^{2}} = \sqrt{k+1}-\sqrt{k} Now, I = k = 1 2016 1 k + 1 + k = 2017 2016 + 2016 2015 . . . . + 2 1 = 2017 1 I=\sum_{k=1}^{2016}\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{2017}-\sqrt{2016}+\sqrt{2016}-\sqrt{2015}....+\sqrt{2}-1=\sqrt{2017}-1 This can be calculate to be approximately 43.9 43.9 , which the ceiling function makes 44 44 .

I'm sorry but I dont't know where to ask this: Is the maths written Mathjax?

Ian Limarta - 4 years, 11 months ago

Log in to reply

It is written in LaTex. Here is a useful LaTex guide. To practice Latex you can use this Google Chrome extension.

Armain Labeeb - 4 years, 11 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...