JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Medium

1 of 1

Analyzing +4 | -1

\left(3+x^{2008}+x^{2009}\right)^{2010}=a_0+a_1 x+a_2 x^2

+\ldots+a_n x^n

, the value of

a_0-\frac{a_1}{2}-\frac{a_2}{2}+a_3-\frac{a_4}{2}-\frac{a_5}{2}+a_6-\ldots

View answer and explanation

Correct option: B

2^{2010}

Option B (

2^{2010}

) is correct because substituting complex cube roots of unity into the given polynomial and summing the results leads to the desired expression.

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