JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Medium

1 of 1

Application +4 | -1

(1+x)^n=\sum_{r=0}^n{ }^n C_r \cdot(x)^r

, then

\left(1+\frac{{ }^n C_1}{{ }^n C_0}\right) \cdot\left(1+\frac{{ }^n C_2}{{ }^n C_1}\right) \cdots\left(1+\frac{{ }^n C_n}{{ }^n C_{n-1}}\right)=

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the expression simplifies to a product of terms that results in this form.

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