IIT JEE Advanced 2008 Paper 2

Since g(x) = logF(x), we get g (x + 1) = log F (x + 1) = log [xF (x)] = log x + log F (x) = log x + g (x) g (x + 1) - g (x) = log x Differentiating on both sides, we get g' (x + 1) - g' (x) = $\frac{1}{x}$ Differentiating again, we get g" (x + 1) - g" (x) - $\frac{1}{x^2}$ Substituting x = x - $\frac{1}{2}$ , we get g" $\left(x+\frac{1}{2}\right)-g''\left(x-\frac{1}{2}\right)$ = $\frac{-(-1)}{\left(x-\frac{1}{2}\right)^2}$ ⇒ $g''\left(x+\frac{1}{2}\right)-g''\left(x-\frac{1}{2}\right)$ = $\frac{-(-4)}{\left(2x-1\right)^2}$ Substituting x = 1, 2, 3, …, N, we get $g''\left(1+\frac{1}{2}\right)-g''\left(1-\frac{1}{2}\right)$ = $\frac{-4}{1}$ $g''\left(2+\frac{1}{2}\right)-g''\left(2-\frac{1}{2}\right)$ = $\frac{-4}{9}$ $g''\left(3+\frac{1}{2}\right)-g''\left(3-\frac{1}{2}\right)$ = $\frac{-4}{25}$ $g''\left(N+\frac{1}{2}\right)-g''\left(N-\frac{1}{2}\right)$ = $\frac{-4}{\left(2N-1\right)^2}$ Hence, we get $g''\left(N+\frac{1}{2}\right)-g''\left(\frac{1}{2}\right)$ = $-4\left[ +\frac{11}{9} + \frac{1}{25} + \dots + \frac{1}{\left(2N-1\right)^2} \right]$