Question

6. Given that the term independent of x in the binomial expansion of (x²+p/x)^6 is 240,

(a) find the value of the positive constant p

Answer 1

By the binomial theorem,

`\displaystyle \left(x^2 + \frac px\right)^6 = \sum_(n=0)^6 \binom 6n \left(x^2\right)^n \left(\frac px\right)^(6-n) = \sum_(n=0)^6 p^(6-n) \binom 6n x^(3n-6)`

where

`\dbinom kn = (k!)/(n!(k-n)!)`

is the so-called binomial coefficient.

The term that's independent of x is the constant term, which occurs when
`3n-6=0`, or
`n=2`. Given that this constant 240, we have

`p^(6-2) \dbinom62 = 240 \implies p^4 = 16 \implies p = \pm2`

but
`p` must be positive, so
`\boxed{p=2}`