Let g(x)=logf(x), where f(x) is a twice differentiable positive function on (0,∞) such that f(x+1)=xf(x). Then, for N=1,2,3,…, g′′(N+21)−g′′(21) is equal to
a. -4 \left\{ 1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{(2N-1)^2} \right\}
b. 4 \left\{ 1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{(2N-1)^2} \right\}
c. -4 \left\{ 1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{(2N+1)^2} \right\}
d. 4 \left\{ 1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{(2N+1)^2} \right\}