# Can you solve it?

The coefficient of x^3 in the expansion of (1+x/2)^n is 5n/12. Find the value of the positive integer n.

### 3 Answers

- husoskiLv 71 month agoFavourite answer
(1 + x/2)^n = ∑ C(n, k) (x/2)^k . . . . sum from k=0 to n

The x^3 term only exists when n>= and then it's

C(n, 3) (x/2)^3 = [n(n-1)(n-2)/6] x^3 / 2^3 = [n(n-1)(n-2) / 48] x^3

You want the coefficient to be 5n/12, so solve:

n(n - 1)(n - 2) / 48 = 5n / 12

The trivial solution n=0 is out, since n>=3 is required above. Multiply by 48/n to get:

(n - 1)(n - 2) = 20

n² - 3n + 2 = 20

n² - 3n - 18 = 0

(n - 6)(n + 3) = 0

You can ignore the n=-3 solution, too, and n=6 is the only feasible result.

Test:

C(n, 3)/2^k = C(6, 3) / 8 = 20 / 8 = 5/2

5n / 12 = 5*6 / 12 = 5/2

Yep...both sides are equal when n=6.

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- Wayne DeguManLv 71 month ago
(1 + x)ⁿ => 1 + nx + n(n - 1)x²/2! + n(n - 1)(n - 2)x³/3! + ...

so, with x => x/2 we have:

(1 + x/2)ⁿ => 1 + n(x/2) + n(n - 1)(x/2)²/2! + n(n - 1)(n - 2)(x/2)³/3! + ...

so, coefficient of x³ => (n/8)(n - 1)(n - 2)/3!

i.e. (n/48)(n - 1)(n - 2) = 5n/12

=> n(n - 1)(n - 2) = 20n

so, (n - 1)(n - 2) = 20

or, n² - 3n - 18 = 0

i.e. (n - 6)(n + 3) = 0

Hence, n = 6

:)>

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- hayharbrLv 71 month ago
Write out the expansion up to the x^3 term:

1^n (x/2)^0 + (nC1) 1^(n-1) (x/2)^1 + (nC2) 1^(n-2) (x/2)^2 + (nC3) 1^(n-3) (x/2)^3 + ...

so the x^3 coefficient will be 1/8 • nC3 = 1/8 n(n-1)(n-2)/6 = n(n-1)(n-2)/48.

n(n-1)(n-2)/48 = 5n/12

Multiply both sides by 12/n and solve

(n-1)(n-2) /4 = 5

(n-1)(n-2) = 20

n^2 - 3n + 2 - 20 = 0

n^2 - 3n - 18 = 0

(n - 6)(n + 3) = 0 and it said n was positive so my guess is n = 6, see what you think

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