If the coefficients of $x$ and $x^2$ in $(1+x)^p(1-x)^q$ are 4 and -5 respectively, then $2 p+3 q$ is equal to

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

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Application +4 | -1

If the coefficients of

x

and

x^2

(1+x)^p(1-x)^q

are 4 and -5 respectively, then

2 p+3 q

is equal to

View answer and explanation

Correct option: D

Option D (63) is correct because by expanding the expression and equating the coefficients of x and x², we can solve for p and q, and then calculate 2p + 3q.

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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