JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

Let

m

and

n

be the coefficients of seventh and thirteenth terms respectively in the expansion of

\left(\frac{1}{3} x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{2}{3}}}\right)^{18}

. Then

\left(\frac{n}{m}\right)^{\frac{1}{3}}

View answer and explanation

Correct option: C

\frac{9}{4}

Option C (

\frac{9}{4}

) is correct because it is the simplified ratio of the coefficients of the seventh and thirteenth terms in the binomial expansion, raised to the power of

\frac{1}{3}

More Questions from BINOMIAL THEOREM

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\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

equals If

a

and

d

are two complex numbers, then the sum to

\left( {n + 1} \right)

terms of the following series

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

is The number of different terms in the expansion of

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

is The sum of the coefficient of the binomial expansion of

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

is equal to 6561 . The constant term in the expansion is The coefficient of

{x^5}

in the expansion of

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

is If

\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

, then

n

equals