The ratio of 7th and 5th terms of expansion (square root 2 +1)^n is 2. What is n?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll write the formula of the general term of the expansion:

Tk+1 = C(n,k)*a^(n-k)*b^k

The 7th term is:

T7 = C(n,6)*(sqrt2)^(n-6)*1^6

The 5th term is:

T5 = C(n,4)*(sqrt2)^(n-4)*1^4

T7/T5 = 2

C(n,6)*(sqrt2)^(n-6)/C(n,4)*(sqrt2)^(n-4) = 2

C(n,6) = n!/6!(n-6)!

C(n,4) = n!/4!(n-4)!

C(n,6)/C(n,4) = 4!(n-4)!/6!(n-6)!

C(n,6)/C(n,4) = 4!(n-6)!(n-5)(n-4)/4!*5*6(n-6)!

We'll simplify and we'll get:

C(n,6)/C(n,4) = (n-5)(n-4)/5*6

C(n,6)/C(n,4) = (n-5)(n-4)/30

(sqrt2)^(n-6-n+4) = (sqrt2)^-2 = 1/2

(n-5)(n-4)/30*2 = 2

(n-5)(n-4) = 120

We'll remove the brackets:

n^2 - 9n + 20 - 120 = 0

n^2 - 9n - 100 = 0

We'll apply quadratic formula:

n1 = [9+sqrt(81 + 400)]/2

Since n is not a natural number, the equation has no solution.

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