The coefficient of $x^5$ in $\left(1+2 x+3 x^2+\ldots\right)^{\frac{-3}{2}}$ is

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Medium

1 of 1

Application +4 | -1

The coefficient of

x^5

\left(1+2 x+3 x^2+\ldots\right)^{\frac{-3}{2}}

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the given expression simplifies to (1-x)³, and the coefficient of x⁵ in this polynomial is 0.

Practice All Chapter MCQs →

\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

a

d

\left( {n + 1} \right)

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

(1-x)^{201}\left(1+x+x^2\right)^{200}

\left(\frac{1}{x}+2 x\right)^n

{x^5}

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170

n