What A Sum!

Algebra Level 3

${\large \frac{1^2 + 2^2}{1 \cdot 2} + \frac{2^2 + 3^2}{ 2 \cdot 3} + . . . + \frac{1003^2 + 1004^2}{ 1003 \cdot 1004} + \frac{1004^2 + 1005^2}{1004 \cdot 1005}}$

What is the nearest integer in the simplified form of the expression above?

The answer is 2009.

2 solutions

Yellow Tomato
Oct 24, 2015

$\color{#EC7300}{\large \large \large \large \frac{1^2 + 2^2}{1 \cdot 2} + \frac{2^2 + 3^2}{ 2 \cdot 3} + . . . + \frac{1003^2 + 1004^2}{ 1003 \cdot 1004} + \frac{1004^2 + 1005^2}{1004 \cdot 1005} = \ ? }$

Simplify for each value:

$\color{#3D99F6}{ \large \large \large \large \frac{1^2 + 2^2}{1 \cdot 2} + \frac{2^2 + 3^2}{ 2 \cdot 3} + . . . + \frac{1003^2 + 1004^2}{ 1003 \cdot 1004} + \frac{1004^2 + 1005^2}{1004 \cdot 1005}}$

$\color{#3D99F6}{\large \Rightarrow \frac {n^2 + (n+1)^2}{n \cdot (n+1)}}$

$\color{#3D99F6}{\large \Rightarrow \frac{n^2}{n(n + 1)} + \frac{ (n + 1)^2}{n(n + 1)}}$

$\color{#3D99F6}{\large \Rightarrow \frac{ n + 1 - 1}{n + 1} + \frac{ n + 1}{n}}$

$\color{#3D99F6}{ \large \Rightarrow 1 - \frac{1}{n+1} + 1 + \frac{1}{n}}$

$\color{#3D99F6}{\large \Rightarrow \frac{1}{n} - \frac{1}{n + 1} + 2}$ .

Let's examine the pattern:

$\color{#D61F06}{\large (1 - \frac{1}{2} + 2) + (\frac{1}{2} - \frac{1}{3} + 2) . . . (\frac{1}{1003} - \frac{1}{1004} + 2) + (\frac{1}{1004} - \frac{1}{1005} + 2)}$

Cancel out positive and negative terms to leave us with:

$\color{#69047E}{\large 2 \cdot 1004 + 1 - \frac{1}{1005}}$

Simplify:

$\color{#20A900}{ \large 2008 + \frac{1004}{1005} \approx 2008.999 }$

Is that pink? Am I color blind?

Vincent Miller Moral - 5 years, 7 months ago

Same here.

Vishal Yadav - 5 years, 7 months ago

It's blue now. That pink was ghastly.

Andrew Ellinor - 5 years, 7 months ago

Haha pink was ghastly. My bad

Yellow Tomato - 5 years, 7 months ago

@Yellow Tomato – No problem! Just easier for those who are color blind to read now.

Andrew Ellinor - 5 years, 7 months ago

Good answer. Can somebody explain about how to get the 1-1/1005 part? I'm confused.

d k - 5 years, 7 months ago

In the series with the $\frac{1}{n}$ part the 1 is $\frac{1}{1}$ and the $- \frac{1}{1005}$ which doesn't get canceled out

Yellow Tomato - 5 years, 7 months ago

You may check the Internet about telescoping series.
1/n - 1/(n+1)

For n=1:
1/1 - 1/2

For n=2:
1/2 - 1/3

For n=3:
1/3 - 1/4

.
.
.

For n=1003:
1/1003 - 1/1004

For n=1004:
1/1004 - 1/1005

If you add all the results, 1/2, -1/2, 1/3, -1/3,...1/1004, -1/1004 will just cancel

Vincent Miller Moral - 5 years, 7 months ago

same way,but i answered first 2008, then 2009. perhaps the question could be worded better. the integer of might mean integer part not nearest integer.

Aareyan Manzoor - 5 years, 7 months ago

Is 2008.999 an integer as per question? I think the question should be worded as "What is the nearest integer in the most simplified form of the expression above?"

Devendra Kumar Singh - 5 years, 7 months ago

Nice problem and solution. Perhaps rather than "most simplified form", (which I found confusing), it would be less ambiguous to simply ask for the nearest integer.

Brian Charlesworth - 5 years, 7 months ago

Zakir Dakua
Nov 4, 2015

The series can be rewritten as $\frac { n^{2}+(n+1)^{2} } {n(n+1)} = \frac {2n^{2}+2n+1} {n^2+n}$

Which is equivalent to $2 + \frac {1}{n^{2}+n}= 2 + \frac{1}{n} - \frac{1}{n+1}$

So we can rewrite it as 2+ $\frac{1}{1} - \frac{1}{2}$ + 2 + $\frac{1}{2} - \frac{1}{3}$ + ... + 2 + $\frac{1}{1004} - \frac{1}{1005}$

= $2 \times 1004$ + $\frac{1}{1} - \frac{1}{1005}$

= 2009 - $\frac{1}{1005}$ $\approx 2009$

What A Sum!

The answer is 2009.

2 solutions

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