Assuming $x$ to be so small that $x^2$ and higher powers of $x$ can be neglected, then $\frac{{\left(1+\frac{3}{4}x\right)}^{-4}{\left(16-3x\right)}^{1/2}}{{(8+x)}^{2/3}}$ is approximately equal to

Find the sum to infinite terms of the series:

$1+\frac{1}{3}+\frac{1·3}{3·6}+\frac{1·3·5}{3·6·9}+\dots \dots \dots .\infty$.

If $x=\frac{1.3}{3.6}+\frac{1.3.5}{3.6.9}+\frac{1.3.5.7}{3.6.9.12}+\dots \dots$, then prove that:

$9x^2+24x=11$.

$\frac{1}{a-b}+\frac{1}{a-2b}+\frac{1}{a-3b}+\dots .+\frac{1}{a-nb}=\alpha n+\beta n^2+\gamma n^3$

then the value of $\gamma$ is :