We have,
\(x=\frac{1}{5}+\frac{1 \cdot 3}{5 \cdot 10}+\frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15}+\ldots\)
Comparing expression with,
\((1+y)^n=\frac{1+n y}{1 !}+\frac{n(n-1)}{2 !} y^2+\ldots .\)
we get,
\(n y=\frac{1}{5}, \frac{n(n-1)}{2 \cdot 1} \cdot y^2=\frac{1 \cdot 3}{5 \cdot 10}\)
Solving we get,
\(y=-\frac{2}{5} \text { and } n=-\frac{1}{2}\)
So,
\(\begin{array}{ll}
x=(1+y)^n-1 \\
\Rightarrow & x=\left(1-\frac{2}{5}\right)^{-\frac{1}{2}}-1=\sqrt{\frac{5}{3}}-1
\end{array}\)
So, \(\quad 3 x^2+6 x\)
\(\begin{aligned}
& =3\left(\sqrt{\frac{5}{3}}-1\right)^2+6\left[\left(\sqrt{\frac{5}{3}}\right)-1\right] \\
& =3\left(\frac{5}{3}+1-2 \sqrt{\frac{5}{3}}\right)+6 \sqrt{\frac{5}{3}}-6 \\
& =5+3-6=2
\end{aligned}\)