The coefficients of the (r-1)th, rth, (r+1)th terms in the expansion of [(x+1)^n] is in the ratio 1:3:5. Find n and r.

 ncert cbse chapter 8 binomial theorem exercise 8.2

 10.The coefficients of the (r-1)th, rth, (r+1)th  terms in the expansion of [(x+1)^n] is in the ratio 1:3:5. Find n and r.

 

T(r+1) =C(n,r)  [x^(n-r) ] [1^r] = C(n,r)  [x^(n-r) ]

T(r)  = C( n,(r-1) )  [x^(n-(r-1) ) ]

T(r-1)= C( n,(r-2) )  [x^(n-(r-2) ) ]

The coefficients of  (r-1)th, rth, (r+1)th  terms in the expansion of [(x+1)^n] 

are C( n,(r-2) ) , C( n,(r-1) ) , C(n,r)  

  C( n,(r-2) ) : C( n,(r-1) ): C(n,r) = 1 : 3  :  5

 C( n,(r-2) ) / C( n,(r-1) ) = 1/3

[(n!) / {(r-2)! *(n-(r-2))!}] / [(n!) / {(r-1)! *(n-(r-1))!}] =1 / 3

 

[{ (r-1)! * (n-r+1)! } / { (r-2)! * (n-r+2)! } ] = 1 / 3

 

using  

(r-1)! = (r-1) * [(r-2)!]

(n-r+2)! = (n-r+2) *[(n-r+1)!] and cancelling 

[(r-1)  / (n-r+2) ] = 1 / 3

3r-3 = n-r+2

4r - n = 5 -------------(1)

C( n,(r-1) ) / C( n,(r) ) = 3/5

 [(n!) / {(r-1)! *(n-(r-1))!}] / [(n!) / {(r)! *(n-r)!}] = 3/5

{(r)! *(n-r)!} / {(r-1)! * (n-r+1)!} = 3/5

 

use 

r!  = (r) * [(r-1)!]

(n-r+1)! =(n-r+1)* [(n-r)!] and cancelling

[r] / [n-r+1] = 3/5

5r =3n-3r+3

8r-3n =3        -----------------(2)

solving (1) and (2)  using elimination method

r=3

 n=7

 

 

7. Find the middle terms in the expansion of [3 - ((x^3) / 6)]^7

There are (7+1) = 8 terms in the expansion

middle terms are 4th and 5th terms

T(r+1) =C(7,r) [3^(7-r)]*[ [(-x^3) / 6)]^r ]

put r=3

T(4) = C(7,3) [3^(7-3)]*[ [(-x^3) / 6)]^3 ]

 T(4) =35* [3^4]* [  [(-x^3) / 6)]^3 ]

T(4)= [- 105/8] * (x^9)

put r=4 in T(r+1) =C(7,r) [3^(7-r)]*[ [(-x^3) / 6)]^r ]

T(5) =  C(7,4) [3^(7-4)]*[ [(-x^3) / 6)]^4 ]

T(5)  = 35* [3^3] * [ [(-x^3) / 6)]^4 ]

T(5)=[35/48] *(x^12)

middle terms are

[- 105/8] * (x^9) and [35/48] *(x^12)

 

 

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 

 6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4 

solution

 

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

solution  

 

8.find n if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of [(fourth root of 2) + {1/(fourth root of 3)}]^n is (sqrt6):1 

solution

 

exercise 8.2

 

 5. find the 4th term in the expansion of (x-2y)^12

solution

7. Find the middle terms in the expansion of [3 - ((x^3) / 6)]^7

solution

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

solution

 

 10.The coefficients of the (r-1)th, rth, (r+1)th  terms in the expansion of [(x+1)^n] is in the ratio 1:3:5. Find n and r.

solution

 

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