Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Try this! - Binomial theorem and its simple applications - JEE Main

For x\epsilon R,x\neq 1 if

(1+x)2016 + x(1+x)2015 + x2(1+x)2014+........+x2016 =    \sum_{i=0}^{2016}\, a_{i}\: x^{i}then a17 is equals to:

 

 

  • Option 1)

    \frac{2017!}{17! \, 2000!}

  • Option 2)

    \frac{2016!}{17! \, 1999!}

  • Option 3)

    \frac{2017!}{ 2000!}

  • Option 4)

    \frac{2016!}{ 16!}

 

Answers (1)

best_answer

As we have learned

Expression of Binomial Theorem -

\left ( x+a \right )^{n}= ^{n}\! c_{0}x^{n}a^{0}+^{n}c_{1}x^{n-1}a^{1}+^{n}c_{2}x^{n-2}a^{2}x-----^{n}c_{n}x^{0}a^{n}

 

- wherein

for n  +ve integral .

 

It is a G.P with C . R = x / 1+x 

So , S = \frac{(1+x)^{2016} \left ( \frac{x}{1+x}^{2017}-1 \right )}{(\frac{x}{1+x})-1}

= -x ^{2017}+ (1+x)^{2017}\\

 

= ^{2017 } C_0 + ^{2017}C_1 x + ^{2017}C_2 X^2 + .....+ ^{2017}C_{2017}X^{2017}- x ^{2017}

^{2017 }C_0 +^{2017 } C_x + ^{2017}C_2x^2 + .....

For \: \: \: x^{17 } , 9 _{17} = ^{2017}C_1_7

= \frac{(2017)!}{(2000)!(17)!}

 

 

 

 

 

 

 

 

 

 


Option 1)

\frac{2017!}{17! \, 2000!}

Option 2)

\frac{2016!}{17! \, 1999!}

Option 3)

\frac{2017!}{ 2000!}

Option 4)

\frac{2016!}{ 16!}

Posted by

gaurav

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 Result 2024 Live: JEE April results at jeemain.nta.ac.in likely today; cutoff, rank predictor, update
anam-khan

JEE Main 2024 session 2 final answer key out at jeemain.nta.ac.in; steps to check probable scores
anam-khan

JEE Main Result 2024 Live: Download session 2 final answer key at jeemain.nta.ac.in; rank predictor, cutoff

Latest Question