Question

Urgent!

Answer 1

Answer:

${(1 - 2x)}^{p} = 1 + ( - 2x)p + \frac{1}{2!} p(p - 1) {( - 2x)}^{2} + ... \\ \frac{4}{2!} p(p - 1) = 40 \\ p(p - 1) = 20 \\ {p}^{2} - p = 20 \\ {p}^{2} - p - 20 = 0\\ ( p - 5)(p + 4) = 0\\ p = 5 \: or \: p = - 4 \\ \because \: p > 0 \\ \therefore \: p = 5$ Apply the binomial expansion...