Question

The coefficient of x in the expansion of (x+1/ax^2)^7 is 7/3. Find the possible values of a.

Answer 1

is it \[(x + \frac{1}{a}x^{2})^{7}\]

Answer 2

If it is, then there is no x in the expansion, there is x^7 all the way to x^14

Answer 3

no it's \[(x + 1/(ax ^{2}) )^{7}\]

Answer 4

ok, after expansion the x coefficient is (21/a^2)
\[\frac{7}{3} = \frac{21}{a^{2}}\]
\[a^{2} = \frac{21*3}{7} = 9\]
\[a = \pm 3\]

Answer 5

ok I expanded the binomial which term do I use?

Answer 6

the term that has x to the power of 1
should be 21/a^2

you need to simplify the terms first

Answer 7

The term with x to the first power is 7x(1/(ax^2))^6 right?

Answer 8

No, if you simplify that term, you get --> 7/(a^6*x^11)
Use term 21x^5(1/(ax^2))^2 = (21x^5)/(a^2*x^4) = 21x/a^2