Question

The coefficients of three consecutive terms in the expression of (1+x)n are 165, 330 and 462. then the value of n is

Answer 1

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Here is your answer...✌

$Let \: the \: three \: consecutive \: terms \: in \: \\ the \: expression \: of\:{ (1 + x) }^{n} \: are \\ ^{n} C _{r} \: ,\: ^{n} C _{r + 1} \: \: and \: \: ^{n} C _{r + 2}$

Therefore,

$^{n}C_{r} = 165 \: \: \: \: \: \: \: \: \: \: . ....(1)\\ \\ ^{n}C_{r + 1} = 330 \: \: \: \: \: \: .....(2)\\ \\ ^{n}C_{r + 2} = 462 \: \: \: \: \: \: .....(3)$

On dividing (2) by (1)
We get,

$\frac{^{n}C_{r + 1}}{^{n}C_{r}} = \frac{330}{165} \\ \\ \frac{ \frac{n!}{(n - r - 1)! \times (r + 1)!} }{ \frac{n!}{(n - r)! \times r!} } = 2 \\ \\ \frac{(n - r)! \times r!}{(n - r - 1)! \times (r + 1)!} = 2 \\ \\ \frac{(n - r) \times (n - r - 1)! \times r!}{(n - r - 1)! \times (r + 1) \times r!} = 2 \\ \\ \frac{n - r}{r + 1} = 2 \\ \\ n - r = 2(r + 1)\\ \\ n = 2r + 2 + r \\ \\ n = 3r + 2 \: \: \: \: \: \: \: \: \: .....(4)$

On dividing (3) by (2)
We get,

$\frac{^{n}C_{r + 2}}{^{n}C_{r + 1}} = \frac{462}{330} \\ \\ \frac{ \frac{n!}{(n - r - 2)! \times (r + 2)!} }{ \frac{n!}{(n - r - 1)! \times (r + 1)!} } = \frac{7}{5} \\ \\ \frac{(n - r - 1)! \times (r + 1)!}{(n - r - 2)! \times (r + 2)!} = \frac{7}{5} \\ \\ \frac{(n - r - 1) \times (n - r - 2)! \times (r + 1)!}{(n - r - 2)! \times (r + 2) \times (r + 1)!} = \frac{7}{5} \\ \\ \frac{n - r - 1}{r + 2} = \frac{7}{5} \\ \\ 5(n - r - 1)= 7(r + 2)\\ \\ 5n -5r - 5 = 7r + 14\\ \\ 5n - 5r - 7r = 14 + 5 \\ \\ 5(3r + 2) - 12r = 19 \: \: \: \: \: \: \: \: | \: from \: (4) \: | \\ \\ 15r + 10 - 12r = 19 \\ \\ 3r = 19 - 10 \\ \\3 r = 9 \\ \\ = > r = 3 \\ \\ putting \: \: in \: \: (4) \\ we \: \: get \\ \\ n = 3(3) + 2 = 9 + 2\\ \\ = > n = 11$

Answer 2

Answer:

The value of n is 11.

Step-by-step explanation:

Given : The coefficients of three consecutive terms in the expression of $(1+x)^n$ are 165, 330 and 462.

To find : The value of n ?

Solution :

Let the three consecutive terms in the expression of $(1+x)^n$ are

$^{n} C _{r} \: ,\: ^{n} C _{r + 1} \: \: and \: \: ^{n} C _{r + 2}$

 $^{n}C_{r} = 165 \: \: \: \: \: \: \: \: \: \: . ....(1)\\ \\ ^{n}C_{r + 1} = 330 \: \: \: \: \: \: .....(2)\\ \\ ^{n}C_{r + 2} = 462 \: \: \: \: \: \: .....(3)$

Divide (2) by (1),

$\frac{^{n}C_{r + 1}}{^{n}C_{r}} = \frac{330}{165} \\ \\ \frac{ \frac{n!}{(n - r - 1)! \times (r + 1)!} }{ \frac{n!}{(n - r)! \times r!} } = 2 \\ \\ \frac{(n - r)! \times r!}{(n - r - 1)! \times (r + 1)!} = 2 \\ \\ \frac{(n - r) \times (n - r - 1)! \times r!}{(n - r - 1)! \times (r + 1) \times r!} = 2 \\ \\ \frac{n - r}{r + 1} = 2 \\ \\ n - r = 2(r + 1)\\ \\ n = 2r + 2 + r \\ \\ n = 3r + 2 \: \: \: \: \: \: \: \: \: .....(4)$

On dividing (3) by (2),

$\frac{^{n}C_{r + 2}}{^{n}C_{r + 1}} = \frac{462}{330} \\ \\ \frac{ \frac{n!}{(n - r - 2)! \times (r + 2)!} }{ \frac{n!}{(n - r - 1)! \times (r + 1)!} } = \frac{7}{5} \\ \\ \frac{(n - r - 1)! \times (r + 1)!}{(n - r - 2)! \times (r + 2)!} = \frac{7}{5} \\ \\ \frac{(n - r - 1) \times (n - r - 2)! \times (r + 1)!}{(n - r - 2)! \times (r + 2) \times (r + 1)!} = \frac{7}{5}$

 $\frac{n - r - 1}{r + 2} = \frac{7}{5} \\ \\ 5(n - r - 1)= 7(r + 2)\\ \\ 5n -5r - 5 = 7r + 14\\ \\ 5n - 5r - 7r = 14 + 5 \\ \\ 5(3r + 2) - 12r = 19 \: \: \: \: \: \: \: \: | \: from \: (4) \: | \\ \\ 15r + 10 - 12r = 19 \\ \\ 3r = 19 - 10 \\ \\3 r = 9 \\ \\ = > r = 3$

Substitute the value in (4),

$n = 3(3) + 2 = 9 + 2\\ \\ = > n = 11$

Therefore, The value of n is 11.