JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

The integer just greater than

(\sqrt{3}+1)^{2 m}

contains

View answer and explanation

Correct option: B

2^{m+1}

as a factor

Option B (

2^{{m+1}}

as a factor) is correct because by expressing

(\sqrt{{3}}+1)^{{2 m}}

I+F

and using its conjugate, we can show that

2^{{m+1}}

is a factor of the integer just greater than

(\sqrt{{3}}+1)^{{2 m}}

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