Question

For the given expansion$(1+2x+3x^2+....{)}^{-3/2},(|x|<1)$

• A. sum of all coefficient will be $0$
• B. coefficient of $x^5$ will be $0$
• C. sum of all coefficient will be $8$
• D. coefficient of $x^2$ will be $3$

Answer 1

Answer: (A), (B), (D)

The correct options are
A sum of all coefficient will be $0$
B coefficient of $x^5$ will be $0$
D coefficient of $x^2$ will be $3$

As we know that
$(1-x{)}^{-2}=1+2x+\frac{2\cdot 3}{2}{x}^2+\frac{2\cdot 3\cdot 4}{2\cdot 3}{x}^3+\cdots =1+2x+3x^2+4x^3+\cdots$

$\therefore (1+2x+3x^2+....{)}^{-3/2}=[(1-x{)}^{-2}{]}^{-3/2}$
$=(1-x{)}^3=1-3x+3x^2-x^3$