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  

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