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Telescoping Surd Sum to 2008 + 2008/2009 | JEE

JEE Maths question with a full step-by-step solution.

Question
The value of ∑n=12008n4+2n3+3n2+2n+1n(n+1)\displaystyle\sum_{n=1}^{2008}\frac{\sqrt{n^4+2n^3+3n^2+2n+1}}{n(n+1)} is
A2008+200820092008+\frac{2008}{2009}correct
B2007+200820092007+\frac{2008}{2009}
C2008+200720092008+\frac{2007}{2009}
D2008+200720082008+\frac{2007}{2008}
Solution
Step 1: Pull n2n^2 out of the radicand:
n4+2n3+3n2+2n+1=n2(n2+2n+3+2n+1n2)=n2(n+1n+1)2.n^4+2n^3+3n^2+2n+1=n^2\left(n^2+2n+3+\frac{2}{n}+\frac{1}{n^2}\right)=n^2\left(n+\frac1n+1\right)^2.
 Step 2: Take the square root:
n4+2n3+3n2+2n+1=n(n+1n+1)=n2+n+1.\sqrt{n^4+2n^3+3n^2+2n+1}=n\left(n+\frac1n+1\right)=n^2+n+1.
 Step 3: The general term becomes
n2+n+1n(n+1)=1+1n(n+1)=1+1n−1n+1.\frac{n^2+n+1}{n(n+1)}=1+\frac{1}{n(n+1)}=1+\frac1n-\frac{1}{n+1}.
 Step 4: Sum from n=1n=1 to 20082008 — the constant gives 20082008, the rest telescopes:
∑n=12008=2008+(1−12009)=2008+20082009.\sum_{n=1}^{2008}=2008+\left(1-\frac{1}{2009}\right)=2008+\frac{2008}{2009}.
 Correct answer: (1)
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