JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

{\left( {1 + x} \right)^{2n}} = {a_0} + {a_1}\;x + {a_2}\;{x^2} + \ldots + {a_{2n}}{x^{2n}},

then

{\left( {{a_0} - {a_2} + {a_4} - {a_6} + \ldots - {a_{2n}}} \right)^2} + {\left( {{a_1} - {a_3} + {a_5} - {a_7} + \ldots + {a_{2n - 1}}} \right)^2}

is equal to

View answer and explanation

Correct option: B

{4^n}

Option B (

{4^n}

) is correct because by substituting

x = i

and

x = -i

into the given expression and using complex number properties, we can simplify the expression to

{4^n}

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