JEE Main 2026 — 6 April, Evening Shift — Question 26

First, simplify the expression inside the square root. Factorize: $n^4+2n^3+3n^2+2n+1 = (n^2+n+1)^2$. Thus, $\sqrt{n^4+2n^3+3n^2+2n+1} = n^2+n+1$. The term becomes: $\frac{n^2+n+1}{n(n+1)} = \frac{n(n+1)+1}{n(n+1)} = 1 + \frac{1}{n(n+1)}$. Now, $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ (partial fraction decomposition). So the sum is: $\sum_{n=1}^{2008} \left[1 + \left(\frac{1}{n} - \frac{1}{n+1}\right)\right]$. This equals $\sum_{n=1}^{2008} 1 + \sum_{n=1}^{2008} \left(\frac{1}{n} - \frac{1}{n+1}\right)$. The first sum is 2008. The second sum telescopes: $\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{2008} - \frac{1}{2009}\right) = 1 - \frac{1}{2009}$. Thus, total sum = $2008 + 1 - \frac{1}{2009} = 2009 - \frac{1}{2009}$. But note: $2009 - \frac{1}{2009} = 2008 + \frac{2008}{2009}$.