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JEE Advanced 2013Paper 1
Exam
JEE Advanced
Year
2013
Paper
Paper 1
Subject
Mathematics
Chapter
Binomial Theorem

JEE Advanced 2013 · Paper 1 · Mathematics · Binomial Theorem

The coefficients of three consecutive terms of& (1 + x) ^ n + 5 are...

The coefficients of three consecutive terms of& (1+x)n+5{\left(1 + \text{x}\right)}^{\text{n} + 5} are in the ratio 5 : 10 : 14\text{5 : 10 : 14}. Then n=\text{n} =

Correct answer

6

Step-by-step solution

n+5Cr−1:n+5Cr:n+5Cr+1=5 : 10 :14\text{n} + 5{\text{C}}_{\text{r} - 1}\text{:}\text{n} + 5{\text{C}}_{\text{r}}\text{:}\text{n} + 5{\text{C}}_{\text{r} + 1} = \text{5 : 10 :14}

⇒(n+5)!(r−1)!(n−r+6)!:(n+5)!(r)!(n−r+5)!:(n+5)!(r+1)!(n−r+4)!=5 : 10 :14⇒ \dfrac{\left(n + 5\right) !}{\left(r - 1\right) ! \left(n - r + 6\right) !}: \dfrac{\left(n + 5\right) !}{\left(r\right) ! \left(n - r + 5\right) !}: \dfrac{\left(n + 5\right) !}{\left(r + 1\right) ! \left(n - r + 4\right) !} = \text{5 : 10 :14}

⇒1(r−1)!(n−r+6)!:1(r)!(n−r+5)!:1(r+1)!(n−r+4)!=5 : 10 :14⇒ \dfrac{1}{\left(r - 1\right) ! \left(n - r + 6\right) !}: \dfrac{1}{\left(r\right) ! \left(n - r + 5\right) !}: \dfrac{1}{\left(r + 1\right) ! \left(n - r + 4\right) !} = \text{5 : 10 :14}

⇒(r+1)r(n−r+6)!:r+1(n−r+5)!:1(n−r+4)!=5 : 10 :14⇒ \dfrac{\left(r + 1\right)r}{\left(n - r + 6\right) !}: \dfrac{r + 1}{\left(n - r + 5\right) !}: \dfrac{1}{\left(n - r + 4\right) !} = \text{5 : 10 :14}

⇒(r+1)r:(r+1)(n−r+6):(n−r+6)(n−r+5)=5 : 10 :14⇒ \left(r + 1\right)r: \left(r + 1\right)\left(n - r + 6\right): \left(n - r + 6\right)\left(n - r + 5\right) = \text{5 : 10 :14}

⇒(r+1)(n−r+6)(r+1)r=105and(n−r+6)(n−r+5)(r+1)(n−r+6)=1410⇒ \dfrac{\left(r + 1\right)\left(n - r + 6\right)}{\left(r + 1\right)r} = \dfrac{10}{5}\text{and}\dfrac{\left(n - r + 6\right)\left(n - r + 5\right)}{\left(r + 1\right)\left(n - r + 6\right)} = \dfrac{14}{10}

⇒(n−r+6)r=2and(n−r+5)(r+1)=75⇒ \dfrac{\left(n - r + 6\right)}{r} = 2\text{and}\dfrac{\left(n - r + 5\right)}{\left(r + 1\right)} = \dfrac{7}{5}

⇒n=3r−6andn=12r−185⇒ n = 3r - 6\text{and}\text{n} = \dfrac{12r - 18}{5}

⇒3r−6=12r−185=n⇒ 3r - 6 = \dfrac{12r - 18}{5} = n

⇒15r−30=12r−18⇒ 15r - 30 = 12r - 18

⇒r=4⇒ r = 4

⇒n=6⇒ n = 6

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