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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

please find the closest integer to S

$S=\sum_{a=10}^{2011}\sqrt{1+\dfrac{a^2+(a+1)^2}{(a(a+1))^2}}$

$S=\sum_{a=10}^{2011}\sqrt{1+\dfrac{a^2+(a+1)^2}{(a(a+1))^2}}$

- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

please find the closest integer to S

$S=\sum_{a=10}^{2011}\sqrt{1+\dfrac{a^2+(a+1)^2}{(a(a+1))^2}}$

$S=\sum_{a=10}^{2011}\sqrt{1+\dfrac{a^2+(a+1)^2}{(a(a+1))^2}}$

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- #2

- Feb 7, 2012

- 2,785

$\sqrt{1+\dfrac{a^2+(a+1)^2}{(a(a+1))^2}}$ looks a bit daunting, so why not get out your calculator and see what the first few terms of that sum look like? Putting $a=10$, we get $$\sqrt{1+\dfrac{100+121}{110^2}} = \sqrt{1+\frac{221}{12100}} = \sqrt{\frac{12321}{12100}} = \frac{111}{110}.$$ That's interesting, $12321$ turns out to be a perfect square! Is that just a coincidence? Try the next term, $a=11$. Then we get $$\sqrt{1+\dfrac{121+144}{132^2}} = \sqrt{1+\frac{265}{17424}} = \sqrt{\frac{17689}{17424}} = \frac{133}{132}.$$ No, that can't be a coincidence. What's more, in each case the numerator is just $1$ more than the denominator, which is fairly clearly equal to $a(a+1).$ It must be true that $$\sqrt{1+\dfrac{a^2+(a+1)^2}{(a(a+1))^2}} = \frac{a^2+a+1}{a(a+1)} = 1+ \frac1a - \frac1{a+1}.$$ So the first thing to do is to verify that fact. Then use it to get a good expression for the sum of the series.please find the closest integer to S

$S=\sum_{a=10}^{2011}\sqrt{1+\dfrac{a^2+(a+1)^2}{(a(a+1))^2}}$

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- #3

- Jan 25, 2013

- 1,225

yes ,you got it

S=$2002+\dfrac{1}{10}-\dfrac{1}{2012}$

and the integer closest to S is 2002

S=$2002+\dfrac{1}{10}-\dfrac{1}{2012}$

and the integer closest to S is 2002