Question

Let $g\left(x\right)=\log \left(f\right(x\left)\right)$ where $f\left(x\right)$ is a twice differentiable positive function on $(0,\infty )$ such that $f(x+1)=xf\left(x\right)$ . Then, for $N=1,2,3,\dots g^{\prime \prime }(N+\frac{1}{2})-g^{\prime \prime }\left(\frac{1}{2}\right)=$

• A. $-4(1+\frac{1}{9}+\frac{1}{25}+\dots +\frac{1}{(2N-1{)}^2})$
• B. $4(1+\frac{1}{9}+\frac{1}{25}+\dots +\frac{1}{(2N-1{)}^2})$
• C. $-4(1+\frac{1}{9}+\frac{1}{25}+\dots +\frac{1}{(2N+1{)}^2})$
• D. $4(1+\frac{1}{9}+\frac{1}{25}+\dots +\frac{1}{(2N+1{)}^2})$

Answer 1

The correct option is A $-4(1+\frac{1}{9}+\frac{1}{25}+\dots +\frac{1}{(2N-1{)}^2})$

$g(x+l)=\log \left(f\right(x+1\left)\right)=logx+\log \left(f\right(x\left)\right)$
$=logx+g\left(x\right)$
$\Rightarrow g(x+1)-g\left(x\right)=logx$
$\Rightarrow g^{\prime \prime }(x+1)-g^{\prime \prime }\left(x\right)=-\frac{1}{x^2}$
$g^{\prime \prime }(1+\frac{1}{2})-g^{\prime \prime }\left(\frac{1}{2}\right)=-4$
$g^{\prime \prime }(2+\frac{1}{2})-g^{\prime \prime }(1+\frac{1}{2})=-\frac{4}{9}$
$g^{\prime \prime }(N+\frac{1}{2})-g^{\prime }(N+\frac{1}{2})=-\frac{4}{(2N-1{)}^2}$
Summing up all terms
Hence, $g^{\prime \prime }(N+\frac{1}{2})-g^{\prime \prime }\left(\frac{1}{2}\right)=-4(1+\frac{1}{9}+\cdots +\frac{1}{(2N-1{)}^2})$