Question

Find the sum of the infinite binomial series$1+\frac{1}{3}+\frac{1}{3}\cdot \frac{3}{6}+\frac{1}{3}\cdot \frac{3}{6}\cdot \frac{5}{9}+...$

• A. $3$
• B. $2$
• C. $\sqrt{3}$
• D. $\sqrt{5}$

Answer 1

The correct option is C $\sqrt{3}$

For a binomial expansion, we can write as
We can write as
$1+nx+\frac{n(n-1)}{2}{x}^2...$
Comparing coefficients, we get
$nx=\frac{1}{3}$ ...(i)
$\frac{n(n-1)}{2}{x}^2=\frac{1}{6}$
$x^2{n}^2-n{x}^2=\frac{1}{3}$
$n^2{x}^2-n{x}^2=nx$ ...from(i)
$nx(nx-x-1)=0$
$\frac{1}{3}(\frac{1}{3}-(x+1\left)\right)=0$
$x+1=\frac{1}{3}$
$x=\frac{-2}{3}$ ...(ii)
Therefore $n=\frac{-1}{2}$
Therefore the expression will be
$(1+x{)}^n$
$=(1-\frac{2}{3}{)}^{-\frac{1}{2}}$
$=\sqrt{3}$