Easy Hyperbolic Summation

look at the summation $\sum_{n=1}^\infty x^{2n-2} =\dfrac{1}{1-x^2}$ integrating both side we have $\sum_{n=1}^\infty \dfrac{x^{2n-1}}{2n-1}=\dfrac{\ln\left(\left|\dfrac{1+x}{1-x}\right|\right)}{2}$ divide both sides with x and put $x=\dfrac{1}{\sqrt{2}}$ . $\sum_{n=1}^\infty \dfrac{1}{(2n-1)2^{n-1}}=\dfrac{\ln\left(\left|\dfrac{1+\dfrac{1}{\sqrt{2}}}{1-\dfrac{1}{\sqrt{2}}}\right|\right)}{\sqrt{2}}=\dfrac{2}{\sqrt{2}} \ln(1+\sqrt{2})=\sqrt{2}\ln(1+\sqrt{2})$ we know $arcsinh(z)=\ln(z+\sqrt{z^2+1})$ . at z =1, $arcsinh(1)=\ln(1+\sqrt{2})$ . so,the expression equals $\sum_{n=1}^\infty \dfrac{1}{(2n-1)2^{n-1}}=\sqrt{2}arcsinh(1)$ and $100(2+1)=\boxed{300}$