Question

If $(1+x{)}^n=1+\frac{3}{4}+\frac{3}{4}.\frac{5}{8}+\frac{3}{4}.\frac{5}{8}.\frac{7}{12}+...\infty$ then the set of values of $x$ and $n$ are respectively

• A. $(\frac{1}{2},-\frac{1}{2})$
• B. $(-\frac{1}{2},-\frac{3}{2})$
• C. $(\frac{1}{2},\frac{3}{2})$
• D. None of these

Answer 1

The correct option is B $(-\frac{1}{2},-\frac{3}{2})$

$(1+x{)}^n$
$=1+nx+\frac{n(n-1)}{2}{x}^2...\infty$
Now $nx=\frac{3}{4}$
and $n(n-1)x^2=\frac{15}{16}$
$nx(nx-x)=\frac{15}{16}$
$\frac{3}{4}-x=\frac{5}{4}$ ...($nx=\frac{3}{4}$)
$x=\frac{-1}{2}$
Therefore $n=\frac{-3}{2}$