JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

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Application +4 | -1

{\left( {1 + x - 2\;{x^2}} \right)^6} = 1 + {C_1}x + {C_2}{x^2} + {C_3}{x^3} + \ldots + {C_{12}}{x^{12}},

then the value of

{C_2} + {C_4} + {C_6} + \ldots + {C_{12}},

View answer and explanation

Correct option: C

Option C (31) is correct because it is the sum of the even-indexed coefficients

{C_2} + {C_4} + {C_6} + \ldots + {C_{12}}

obtained by substituting

x = 1

and

x = -1

into the given equation and solving.

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