The coefficient of x in the expansion of [x+(1/ax^2)]^7 is 7/3. Find the possible values of a.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Homework Help: Binomial Expansion

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