JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Hard

1 of 1

Application +4 | -1

In the expansion of the following expression

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

, the coefficient of

{x^4}\left( {0 \le k \le n} \right)

View answer and explanation

Correct option: A

{\;^{n + 1}}{C_{k + 1}}

Option A (

{\;^{n + 1}}{C_{k + 1}}

) is correct because the sum of the geometric progression can be simplified, and the coefficient of x^k is then found using the binomial theorem.

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