Question

If $|x|<1$, then $\frac{d}{dx}[1+\frac{p}{q}x+\frac{p(p+q)}{2!}{\left(\frac{x}{q}\right)}^2+\frac{p(p+q)(p+2q)}{3!}{\left(\frac{x}{q}\right)}^3....\infty ]=$

• A. $\frac{p}{q(1-x{)}^{\frac{p}{q}+1}}$
• B. $\frac{p}{q(1-x{)}^{\frac{p}{q}}}$
• C. $(1-x{)}^{-pq-1}$
• D. $(1-x{)}^{pq+1}$

Answer 1

The correct option is A $\frac{p}{q(1-x{)}^{\frac{p}{q}+1}}$

$\frac{d}{dx}\{1+\frac{px}{q}+\frac{p(p+q)}{2!}\frac{x^2}{q^2}+\dots \}=\frac{p}{q}+\frac{p(p+q)}{2!}\frac{2x}{q^2}+\frac{p(p+q)(p+2q)}{3!}\frac{3x^2}{q^3}+\dots$

$=\frac{p}{q}(1+\frac{(p+q)}{1!}\frac{x}{q}+\frac{(p+q)(p+2q)}{2!}\frac{x^2}{q^2}+\dots )$

Now, consider $(1-a{)}^n=1-na+\frac{n(n-1)}{2!}{x}^2+\dots$

Put $a=x$ and $n=-(1+\frac{p}{q})$

$\therefore (1-x{)}^{-(1+\frac{p}{q})}=1+\frac{(p+q)}{q}x+[-\frac{(p+q)}{q}(-\frac{(p+q)}{q}-1)]\frac{x^2}{2!}\dots$

$=1+\frac{(p+q)}{q}x+\left[\frac{(p+q)}{q}\left(\frac{(p+q+q)}{q}\right)\right]\frac{x^2}{2!}\dots$

$=1+\frac{(p+q)}{q}x+\frac{(p+q)(p+2q)}{q^2}\frac{x^2}{2!}\dots$

Hence,

$\frac{d}{dx}\{1+\frac{px}{q}+\frac{p(p+q)}{2!}\frac{x^2}{q^2}+\dots \}=\frac{p}{q}(1-x{)}^{-(1+\frac{p}{q})}$

This is the required answer.