Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Find the total number of terms in the expansion of <img alt="\small \left(x^3+1+\frac{1}{x^3}\right)^{100}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Csmall%20%5Cleft%28x%5E3&

Find the total number of terms in the expansion of \small \left(x^3+1+\frac{1}{x^3}\right)^{100}

Option: 1

201


Option: 2

200


Option: 3

300


Option: 4

301


Answers (1)

best_answer

\left[1+\left(x^3+\frac{1}{x^3}\right)\right]^{100}=1+{ }^{100} C_1\left(x^3+\frac{1}{x^3}\right)^1+{ }^{100} C_2\left(x^3+\frac{1}{x^3}\right)^2+\ldots \ldots+{ }^{100} C_{100}\left(x^3+\frac{1}{x^3}\right)^{100}

expansion contain constant term and terms obtained by combination of  x^3 \text { and } \frac{1}{x^3}

 will get converted into a terms involving x^3 \text { and } \frac{1}{x^3}

\begin{aligned} & \Rightarrow \text { constant }+\alpha_1 x^3+\alpha_2 x^6+\ldots \ldots \ldots+\alpha_{100}\left(x^3\right)^n+\beta_1 \frac{1}{x^3}+\beta_2\left(\frac{1}{x^3}\right)^2+\ldots \ldots \ldots+\beta_{100}\left(\frac{1}{x^3}\right)^{\prime} \\ & \text { Hence total number of terms }=1+100+100=201 . \end{aligned}
 

Posted by

seema garhwal

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 Registration LIVE: NTA JEE form at jeemain.nta.nic.in; what's new this year?
anam-khan

JEE Main 2026: Wider gap in NIT cut-offs for home and other-state quotas; outsiders need better rank
anam-khan

JEE Mains 2026: Use of calculator 'not allowed’ says NTA; cites ‘typographical error' in information bulletin

Latest Question