JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

The constant term in the expansion of

{\left( {1 + x} \right)^{10}}{\left( {1 + \frac{1}{x}} \right)^{12}}

View answer and explanation

Correct option: A

{\;^{22}}{C_{10}}

Option A (

{{\;^{22}}}{C_{10}}

) is correct because the constant term in the expansion of

{\left( {1 + x} \right)^{10}}{\left( {1 + \frac{1}{x}} \right)^{12}}

is the coefficient of

x^{12}

{\left( {1 + x} \right)^{22}}

, which is

{{\;^{22}}}{C_{12}} = {{\;^{22}}}{C_{10}}

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