Recall that from the Binomial Theorem, (a+b)ⁿ expands with the terms:

aⁿ, a^n-1b, a^n-2b², ......, ab^n-1, bⁿ

In words: start with aⁿ and then decrease its power by 1, while increasing the power of b by 1.

Hence (2/x + 3x)⁵ expands into the terms:

(2/x)⁵, (2/x)⁴(3x), (2/x)³(3x)², (2/x)²(3x)³, (2/x)(3x)⁴, (3x)⁵

When we evaluate the x power of each term, we get: -5, -3, -1, 1, 3, 5 respectively.

Hence the answers to (i) and (iii) are 0. The coefficient to non-existant terms are zero.

(ii) x³ is the second last term: (2/x)(3x)⁴

Recall that there is also a binomial coefficient to that term: ⁵C₄ = 5

We can also get the binomial coefficients by using row 5 of Pascal's triangle.

1, 5, 10, 10, 5, 1

Useful for grabbing the coefficients in bulk.

So the relevant term is:

• 5(2/x)(3x)⁴
• (10/x)(81x⁴)
• 810x³

Hence, the coefficient is 810.