Test Index

NCERT Class XI Mathematics - Binomial Theorem - Solutions

© examsnet.com
Question : 5 of 36
Marks: +1, -0
(x+1x)6\left(x+\frac{1}{x}\right)^6
Solution:  
We have, (x+1x)6\left(x+\frac{1}{x}\right)^6
=
(60)x6+(61)x5(1x)+(62)x4(1x)2+(63)x3(1x)3+(64)x2(1x)4+(65)x(1x)5+(66)(1x)6\binom{6}{0}x^6 + \binom{6}{1}x^5\left(\frac{1}{x}\right) + \binom{6}{2}x^4\left(\frac{1}{x}\right)^2 + \binom{6}{3}x^3\left(\frac{1}{x}\right)^3 + \binom{6}{4}x^2\left(\frac{1}{x}\right)^4 + \binom{6}{5}x\left(\frac{1}{x}\right)^5 + \binom{6}{6}\left(\frac{1}{x}\right)^6
= x6+6x4+15x2x^6+6x^4+15x^2 + 20 + 15 (1x2)+6(1x4)+1x6\left(\frac{1}{x^2}\right)+6\left(\frac{1}{x^4}\right)+\frac{1}{x^6}
= x6+6x4+15x2x^6+6x^4+15x^2 + 20 + 15x2+6x4+1x6\frac{15}{x^2}+\frac{6}{x^4}+\frac{1}{x^6}
© examsnet.com
Go to Question:
Prev Question
Next Question