auto ad

Wednesday, August 12, 2020

find an approximate value of (0.99^5) using the first three terms of its expansion

 ncert cbse chapter 8 binomial theorem miscellaneous exercise  

7.find an approximate value of (0.99^5) using the first three terms of its expansion

(1-x)^5 = 1 -C(5,1)x + C(5,2)(x^2)+ higher powers of x

C(5,1)=5

C(5,2)=10

(1-x)^5 = 1 -5x +10(x^2)+ higher powers of x

choose x =0.01

(1-0.01)^5 =1 -5(0.01) +10((0.01)^2)+ higher powers of 0.01

0.99^5 = 1 - 0.05 +0.001 = 0.951 approximately.

exercise 8.2

Q8) Find the  middle terms in the expansion of [(x/3)+9y)]^10

There are (10 +1) = 11 terms in the expansion

middle term is the 6th term

T(r+1) = C(10,r) [(x/3)^(10-r)][(9y)^(r)]

put r=5

T(6) =C(10,5) [(x/3)^(10-5)][(9y)^(5)]

= [252]*[(x^5)/243][59049 (y^5)]

=61236(x^5)(y^5)


chapter 8 binomial theorem miscellaneous exercise

 1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

 

2.  Find a if the coefficients of (x^2)  & (x^3) in the expansion of {(3+ax)^9} are equal 

solution

  3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

 

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

solution 

7.find an approximate value of (0.99^5) using the first three terms of its expansion

solution  

exercise 8.2

Q8) Find the  middle terms in the expansion of [(x/3)+9y)]^10

solution  

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

No comments:

Post a Comment

please leave your comments