Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • <img alt="\small \left({ }^m C_0+{ }^m C_1-{ }^m C_2-{ }^m C_3\right)+\left({ }^m C_4+{ }^m C_5-{ }^m C_6-{ }^m C_7\right)+\ldots=0" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Csmall%20

\small \left({ }^m C_0+{ }^m C_1-{ }^m C_2-{ }^m C_3\right)+\left({ }^m C_4+{ }^m C_5-{ }^m C_6-{ }^m C_7\right)+\ldots=0

if and only if for some positive integer, then m=

Option: 1

4k


Option: 2

4k+1


Option: 3

4k-1


Option: 4

4k+2


Answers (1)

best_answer

\text { Consider, }(\cos \theta-i \sin \theta)^m

\begin{aligned} ={ }^m C_0 \cos ^m \theta-{ }^m C_1 \cos ^{m-1} \theta & (i \sin \theta)+\ldots \\ & +{ }^m C_m(-i \sin \theta)^m \end{aligned}

Also,\ (\cos \theta+i \sin \theta)^m={ }^m C_0 \cos ^m \theta +{ }^m C_1 \cos ^{m-1} \theta(i \sin \theta)+\ldots+{ }^m C_m(i \sin \theta)^m

\text{Adding (i) and (ii), we get} \\ 2 \cos m \theta=2\left[{ }^m C_0 \cos ^m \theta-{ }^m C_2 \cos ^{m-2} \theta \sin ^2 \theta \ldots\right]

\text{Subtracting (i) from (ii), we get} \\ 2 i \sin m \theta=2 i\left[{ }^m C_1 \cos ^{m-1} \theta \sin \theta-{ }^m C_3 \cos ^{m-3} \theta \sin ^3 \theta \ldots\right]

Adding (iii) and (iv), we get

\begin{array}{r} \cos m \theta+\sin m \theta=\left[{ }^m C_0 \cos ^m \theta+{ }^m C_1 \cos ^{m-1} \theta \sin \theta-\right. \\ \left.{ }^m C_2 \cos ^{m-2} \theta \sin ^2 \theta-{ }^m C_3 \cos ^{m-3} \theta \sin ^3 \theta \ldots\right] \\ \Rightarrow \sqrt{2} \sin \left(m \theta+\frac{\pi}{4}\right)=\left[{ }^m C_0 \cos ^m \theta+{ }^m C_1 \cos ^{m-1} \theta \sin \theta\right. \\ \left.-{ }^m C_2 \cos ^{m-2} \theta \sin ^2 \theta-{ }^m C_3 \cos ^{m-3} \theta \sin ^3 \theta \ldots\right] \end{array}

\text { Putting } \theta=\frac{\pi}{4} \text {, we get }

\begin{aligned} \sqrt{2} \sin \left(\frac{(m+1) \pi}{4}\right)= & \frac{1}{2^{m / 2}}\left[\left({ }^m C_0+{ }^m C_1-{ }^m C_2-{ }^m C_3\right)+\right. \\ & +\left({ }^m C_4+{ }^m C_5-{ }^m C_6-{ }^m C_7\right)+\ldots \\ & \left.\left.{ }^{m-3}{ }^m C_{m-2}-{ }^m C_{m-1}-{ }^m C_m\right)\right] \end{aligned}

Hence, m + 1 = 4k, for given quantity to be 0.

\Rightarrow m=4 k-1 \text {, where } k \in N

 

Posted by

shivangi.bhatnagar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow
anam-khan

JEE Main 2026 Registration LIVE: Session 2 exam from April 2 to 9; shift timings, admit card, exam pattern
anam-khan

JEE Main 2026 session 2 registration ends today for BTech, BArch; apply at jeemain.nta.nic.in

Latest Question