Math, asked by chandranshekhar3075, 1 year ago

if three consecutive terms in the expansion of (1+x)^n are in the ratio of 6:33:110. find
n and r.
please don't give me any link. i have already seen them but the way it has been done in the similar query, the answer for the given question doesn't come
the answer comes only when we take the coefficients as nCr-1, nCr, nCr+1 tell me if it is right because in the two different ways , different answers come .
plz fast, exam 2morrow.

Answers

Answered by Anonymous
23
n=12 r=2 i hope there r the values
Attachments:
Answered by amitnrw
0

¹²C₁  , ¹²C₂  , ¹²C₃  are terms in the ratio of 6:33:110

Step-by-step explanation:

(1 + x)ⁿ  = 1 + ⁿC₁x¹  + ⁿC₂x²+..............................+ⁿCₙxⁿ

Three consecutive terms coefficients

ⁿCₐ  : ⁿCₐ₊₁  :  ⁿCₐ₊₂  :::  6 : 33 :  110

=>  ⁿCₐ  = 6K  =>  n!/(a!)(n-a)!  = 6K  => n! = 6K (a!)(n-a)!

ⁿCₐ₊₁   = 33K   => n!/(a+1)!(n-a-1)! = 33K  => n!  = 33K (a+1)!(n-a-1)!

ⁿCₐ₊₂ = 110K  => n!/(a + 2)!(n-a-2)! = 110K  => n! = 110K (a + 2)!(n-a-2)!

6K (a!)(n-a)!  = 33K (a+1)!(n-a-1)!

=> 2  (a!)(n-a)(n-a - 1)! = 11 (a + 1)a! (n-a-1)!

=> 2(n-a) = 11(a + 1)

=> 2n - 2a = 11a + 11

=> 2n = 13a + 11

=> 13a = 2n - 11

33K (a+1)!(n-a-1)!  = 110K (a + 2)!(n-a-2)!

=> 3 (a+1)!(n-a-1)(n-a-2)!  = 10 (a + 2)(a + 1)!(n-a-2)!

=> 3 (n - a - 1) = 10(a + 2)

=> 3n - 3a - 3 = 10a + 20

=>  3n = 13a + 23

=> 13a = 3n - 23

2n - 11 = 3n - 23

=> n = 12

13a = 2n - 11

=> 13a = 2*12 - 11

=> 13a = 13

=> a = 1

this means

¹²C₁  , ¹²C₂  , ¹²C₃  are terms

12  :  66  : 220

6 : 33 : 110

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