JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

1 of 1

Analyzing +4 | -1

Let

{(1 + x + 2{x^2})^{20}} = {a_0} + {a_1}x + {a_2}{x^2} + ..... + {a_{40}}{x^{40}}

, then

{a_1} + {a_3} + {a_5} + .... + {a_{37}}

is equal to

View answer and explanation

Correct option: B

{2^{19}}({2^{20}} - 21)

Option B (

{2^{19}}({2^{20}} - 21)

) is correct because it is derived by substituting x = 1 and x = -1 into the polynomial and then subtracting the resulting equations.

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