JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

Let

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

Then,

\left( {1 + \frac{{{a_1}}}{{{a_0}}}} \right)\left( {1 + \frac{{{a_2}}}{{{a_1}}}} \right) \ldots \left( {1 + \frac{{{a_n}}}{{{a_{n - 1}}}}} \right)\;

is equal to

View answer and explanation

Correct option: B

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

Option B is correct because the given expression simplifies to

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

after applying the binomial coefficient properties and simplifying the product.

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