$x=\frac{3}{4 \cdot 8}+\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}+\ldots \ldots \ldots$
Let,
$$
\begin{aligned}
(1+y)^n=1+n y+ & \frac{n(n-1)}{2 !} y^2+\frac{n(n-1)(n-2)}{3 !} \\
& +\frac{n(n-1)(n-2)(n-3)}{4 !} y^4+\ldots . .
\end{aligned}
$$
If we compare with third term on words, the
$$
\begin{aligned}
\frac{n(n-1)}{2 !} y^2 & =\frac{3}{4.8}, \frac{n(n-1)(n-2)}{n !} \\
y^3 & =\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}
\end{aligned}
$$
and $\frac{n(n-1)(n-2)(n-3)}{4 !} y^4=\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}$
On solving, we are getting. So,
$$
\begin{aligned}
& x=(1+y)^n-1-n y=\left(1-\frac{1}{2}\right)^{-1 / 2}-1-\left(-\frac{1}{2}\right)\left(\frac{-1}{2}\right) \\
& \Rightarrow \sqrt{2}-1-\frac{1}{4}=\sqrt{2}-\frac{5}{4} \\
& \text { Now, } \quad 2 x^2=2\left[2+\frac{25}{16}-\frac{5}{\sqrt{2}}\right]=4+\frac{25}{8}-5 \sqrt{2} \\
& \text { and } \quad 5 x=5 \sqrt{2}-\frac{25}{4} \\
&
\end{aligned}
$$
So,
$$
\begin{aligned}
2 x^2+5 x & =4+\frac{25}{8}-5 \sqrt{2}+5 \sqrt{2}-\frac{25}{4} \\
& =\frac{32+25-50}{8}=\frac{7}{8}
\end{aligned}
$$