JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Hard

1 of 1

Analyzing +4 | -1

The coefficient of

{x^{20}}

in the expansion of

{\left( {1 + {x^2}} \right)^{40}}{\left( {{x^2} + 2 + \frac{1}{{{x^2}}}} \right)^{ - 5}},

View answer and explanation

Correct option: B

{\;^{30}}{C_{25}}

Option D (

{{\;^{30}}C_{25}}

) is correct because simplifying the expression and finding the coefficient of

x^{10}

(1+x^2)^{30}

leads to

{{\;^{30}}C_5}

which is equal to

{{\;^{30}}C_{25}}

More Questions from BINOMIAL THEOREM

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