If $1+x^4+x^5=\sum_\limits {i=0}^5 a_i(1+x)^i$, for all $x$ in ${R}$, then $a_2$ is:

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

1+x^4+x^5=\sum\limits_{{i=0}}^5 a_i(1+x)^i

, for all

x

{R}

, then

a_2

is:

View answer and explanation

Correct option: A

-4

Option A (-4) is correct because by substituting

x

with

x-1

and comparing coefficients, we can find the value of

a_2

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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