- #1
thornluke
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The coefficient of x in the expansion of [x+(1/ax^2)]^7 is 7/3. Find the possible values of a.
1. Rewrite (x + 1/(ax^2))^7 = x^(-14) (x^3 + 1/a)^7.
So, we need to find the coefficient of x^15 from (x^3 + 1/a)^7.
2. Using the Binomial Theorem, we have
(x^3 + 1/a)^7 = Σ(k = 0 to 7) C(7, k) (x^3)^(7 - k) (1/a)^k.
......= Σ(k = 0 to 7) C(7, k) x^(21 - 3k) (1/a)^k.
3. So, we need 21 - 3k = 15 ==> k = 2.
Thus, we have (1/a)^2 = 7/3
==> a = ±√(3/7).
The problem is, I do not understand the steps. Help please?
1. Rewrite (x + 1/(ax^2))^7 = x^(-14) (x^3 + 1/a)^7.
So, we need to find the coefficient of x^15 from (x^3 + 1/a)^7.
2. Using the Binomial Theorem, we have
(x^3 + 1/a)^7 = Σ(k = 0 to 7) C(7, k) (x^3)^(7 - k) (1/a)^k.
......= Σ(k = 0 to 7) C(7, k) x^(21 - 3k) (1/a)^k.
3. So, we need 21 - 3k = 15 ==> k = 2.
Thus, we have (1/a)^2 = 7/3
==> a = ±√(3/7).
The problem is, I do not understand the steps. Help please?