If ${\left( {1 + 2\;x + {x^2}} \right)^n} = \mathop \sum \limits_{r = 0}^{2\;n} a,\;{x^r},$ Then ${a_r} = $

${\left( {{\;^n}{C_r}} \right)^2}$

${\;^n}{C_r}.{\;^n}{C_{r + 1}}$

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

{\left( {1 + 2\;x + {x^2}} \right)^n} = \sum \limits_{r = 0}^{2\;n} a,\;{x^r},

Then

{a_r} =

View answer and explanation

Correct option: C

{\;^{2n}}{C_r}

Option C (

\;^{2n}{C_r}

) is correct because simplifying the expression and applying the binomial theorem leads to this result.

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