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If three successive terms in the expansion of (1+x)na have their coefficients in the ratio 6 : 33 : 110, then n is equal to
- a)10
- b)11
- c)12
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
If three successive terms in the expansion of(1+x)na have their coeffi...
(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
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
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If three successive terms in the expansion of(1+x)na have their coeffi...
To find the value of n in the given expression, we need to analyze the coefficients of three successive terms in the expansion of (1+x)^n. Let's break down the solution into the following steps:
Step 1: Coefficients of Three Successive Terms
Let the three successive terms be (nCr) * 1^(n-r) * x^r, (nCr+1) * 1^(n-r-1) * x^(r+1), and (nCr+2) * 1^(n-r-2) * x^(r+2), where r is the common power of x in the terms.
Step 2: Ratio of Coefficients
Given that the coefficients are in the ratio 6 : 33 : 110, we can write the ratios as:
(nCr) : (nCr+1) : (nCr+2) = 6 : 33 : 110
Step 3: Ratio Calculation
Using the formula for binomial coefficients, we can simplify the ratios and eventually get the value of r.
Step 4: Calculating n
Once we have the value of r, we can substitute it back into the binomial expression and find the value of n.
Therefore, the correct answer is option 'C' as n will be equal to 12 based on the coefficients ratio provided in the question.
Step 1: Coefficients of Three Successive Terms
Let the three successive terms be (nCr) * 1^(n-r) * x^r, (nCr+1) * 1^(n-r-1) * x^(r+1), and (nCr+2) * 1^(n-r-2) * x^(r+2), where r is the common power of x in the terms.
Step 2: Ratio of Coefficients
Given that the coefficients are in the ratio 6 : 33 : 110, we can write the ratios as:
(nCr) : (nCr+1) : (nCr+2) = 6 : 33 : 110
Step 3: Ratio Calculation
Using the formula for binomial coefficients, we can simplify the ratios and eventually get the value of r.
Step 4: Calculating n
Once we have the value of r, we can substitute it back into the binomial expression and find the value of n.
Therefore, the correct answer is option 'C' as n will be equal to 12 based on the coefficients ratio provided in the question.
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If three successive terms in the expansion of(1+x)na have their coefficients in the ratio 6 : 33 : 110, then n is equal toa)10b)11c)12d)none of theseCorrect answer is option 'C'. Can you explain this answer?
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