JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

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

and

{b_r} = 1 + \frac{{{a_r}}}{{{a_r} - 1}}

and

\mathop \Pi \limits_{r = 1}^n {b_r} = \frac{{{{\left( {101} \right)}^{100}}}}{{100!}}

, then

n

View answer and explanation

Correct option: B

100

Option B (100) is correct because by comparing the given product with the binomial expansion, we can deduce that n = 100.

More Questions from BINOMIAL THEOREM

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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

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