JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

1 of 1

Application +4 | -1

If in the expansion of

\left(2^x+\frac{1}{4^x}\right)^n, T_3 / T_2=7

and the sum of the coefficient of

2^{\text {nd }}

and

3^{\text {rd }}

terms is 36 , then the value of

x

View answer and explanation

Correct option: A

-1 / 3

Option A (-1/3) is correct because solving the given conditions

{ }^n C_1+{ }^n C_2=36

and

\frac{{ }^n C_2(2^x)^{n-2} (\frac{{1}}{{4^x}})^2}{{ }^n C_1(2^x)^{n-1} (\frac{{1}}{{4^x}})^1}=7

leads to

x = -1/3

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