Question

Urgent!
when (1-2x)^P is expanded, the coefficient of x^2 is 40. Given that P>0 find the value of P​

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...

Answer 2

The value of P for the given binomial expansion when calculated is; p = 5

How to Solve Binomial Expansions?

We are told that when we expand the expression (1 - 2x)^p, that the coefficient of x² is 40.

The general formula for binomial expansion is;

(x + y)ⁿ = ⁿC₀ xⁿ y⁰ + ⁿC₁ xⁿ⁻¹ y₁ + ⁿC² xⁿ⁻² y₂ + ... + ⁿCₙ x⁰ yⁿ

Applying that general formula to our question, we have;

(-2x + 1)^p = 1 + (-2x)p + [p(p - 1)/2!](-2x)² + ......

Thus, simplifying this gives;

[p(p - 1)/2!](-2x)² = -2p(p - 1)x²

Since the coefficient of x² is 40, then we have;

-2p(p - 1) = 40

p² - p = -20

p² - p + 20 = 0

Using online quadratic equation calculator, we have;

p = 5

Read more about Binomial Expansions at; https://brainly.com/question/13602562

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