If ${\left( {1 + x + {x^2}} \right)^n} = \mathop \sum \limits_{r = 0}^{2n} {a_r}{x^r},\;{\rm{then\;}}{a_1} - 2{a_2} + 3{a_3} \ldots - 2n{a_{2n}}$ is equal to

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

1 of 1

Analyzing +4 | -1

{\left( {1 + x + {x^2}} \right)^n} = \sum \limits_{r = 0}^{2n} {a_r}{x^r},\;\mathrm{then\;}{a_1} - 2{a_2} + 3{a_3} \ldots - 2n{a_{2n}}

is equal to

View answer and explanation

Correct option: B

- n

Option B (-n) is correct because it is derived by differentiating the given expression and substituting x = -1.

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