If $T_{r+1}$ is the term independent of $x$ in $\left(\sqrt{a} x^2+\frac{1}{2 x^3}\right)^{10}$ then $r=$

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

T_{r+1}

is the term independent of

x

\left(\sqrt{a} x^2+\frac{1}{2 x^3}\right)^{10}

then

r=

View answer and explanation

Correct option: A

Option A (2) is correct because there are two solutions for θ in the given interval that satisfy the trigonometric equation.

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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