1oh yeah ,i meant (n+1) only,it was a typo
\sum_{n=1}^{2008}\frac{{\sqrt{n^{4}+2n^{3}+3n^{2}+2n+1}}}{n(n+1)} =
A) 2008 + 2008/2009 B) 2007 + 2008/2009
C) 2008 + 2007/2009 D) 2008 + 2007/2008
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5 Answers
1THe term simplifies to \frac{\sqrt{(n+\frac{1}{n}+1)^2}}{(n+1)} = 1 + \frac{1}{n} - \frac{1}{n+1}
Therefore the sum is 2008 + \frac{2008}{2009}
62I think there is a small mistake in what vivek has written...
but the basic funda is correct....
divide the numerator by n (take it inside the square root.. so it becomes divided by n^2
now express it as (n+1/n)
and then solve!
1yeah... his solution is correct except d denominator in first expression... it must b '(n+1)' but not 'n(n+1)'
