Let $n$ be an odd integer. If $\sin n{\rm{\theta }} = \mathop \sum \limits_{r = 0}^n {b_r}{\sin ^r}{\rm{\theta }}$ for every value of $\theta $, then

${b_0} = 0,\;{b_1} = {n^2} - 3n + 3$

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Hard

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Analyzing +4 | -1

Let

n

be an odd integer. If

\sin n\mathrm{\theta } = \sum \limits_{r = 0}^n {b_r}{\sin ^r}\mathrm{\theta }

for every value of

\theta

, then

View answer and explanation

Correct option: B

{b_0} = 0,\;{b_1} = n

The correct answer is B (

{b_0} = 0,\;{b_1} = n

) because it is derived from the expansion of

\sin n\theta

using trigonometric identities and comparing coefficients.

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