JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

1 of 1

Application +4 | -1

{\left( {1 + x} \right)^n} = \sum \limits_{r = 0}^n {C_r}{x^r}

, then

\left( {1 + \frac{{{C_1}}}{{{C_0}}}} \right)\;\left( {1 + \frac{{{C_2}}}{{{C_1}}}} \right) \ldots \;\left( {1 + \frac{{{C_n}}}{{{C_{n - 1}}}}} \right)

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is the simplified form of the product of the given expression using binomial coefficients.

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