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
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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