MHB What Is the Closest Integer to This Complex Summation?

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The discussion focuses on calculating the closest integer to the summation S defined as S = ∑ from a=10 to 2011 of √(1 + (a² + (a+1)²)/(a(a+1))²). Initial calculations reveal that the terms simplify to a pattern where √(1 + (a² + (a+1)²)/(a(a+1))²) equals (a² + a + 1)/(a(a+1)). This leads to the conclusion that S can be expressed as S = 2002 + 1/10 - 1/2012. After evaluating this expression, the closest integer to S is determined to be 2002. The final result confirms that the closest integer to the complex summation is indeed 2002.
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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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Albert said:
please find the closest integer to S

$S=\sum_{a=10}^{2011}\sqrt{1+\dfrac{a^2+(a+1)^2}{(a(a+1))^2}}$
$\sqrt{1+\dfrac{a^2+(a+1)^2}{(a(a+1))^2}}$ 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.
 
yes ,you got it (Yes)

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

and the integer closest to S is 2002
 
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