Use the Binomial Theorem to find the numerical value of (0.998)^6 correct to five decimal places

Use the Binomial Theorem to find the numerical value of...

Key Concepts

Binomial Theorem

The Binomial Theorem provides a systematic way to expand expressions of the form (a + b)^n into a sum involving binomial coefficients. It is expressed as (a + b)^n = ? from k=0 to n [nCk * a^(n-k) * b^k], where nCk represents the binomial coefficients. This theorem is invaluable for breaking down complex power expressions into simpler terms that can be computed individually.

Numerical Approximation Using Binomial Expansion

For expressions that are close to 1 raised to a power, such as (1 - x)^n with a small x, the Binomial Theorem can be used to approximate the expression by expanding and summing only a few leading terms. This approach can yield high precision, such as computing values to a specified number of decimal places, by considering the magnitude of omitted terms to ensure the required accuracy.

Transcript

00:01 If we want to find the value of 0 .998 to the 6th, we can use a calculator, or we could use the binomial theorem written above.

00:11 We can rewrite 0 .998 to the 6th as a binomial of the form 1 minus 2 to the negative 3rd equals 6, or 1 plus negative 2 to the negative 3rd all to the 6th power.

00:29 In this form, we could see that x in our equation equals 1, a in our equation equals negative 2 to the negative 3rd, and n equals 6.

00:41 We can expand this using the binomial theorem.

00:45 For the first term, j equals 0.

00:48 So we get 6 to 0 times x, which in our case is 1, to the 6 minus 0 power, times a, which in our case is 1, to the 6 minus 0 power, times a, which in our case, case is negative 2 to the negative third to the j which equals zero now anything raised to the first power any now one raised to any power is going to now one raised to any power is going to equal one so going forward we're just going to disregard the term with we're just going to disregard this term so for the second term of our equation, we j equals 2 and we get, so for the second term of our equation, j equals 1, and we get 6 choose 1, we're ignoring the x term because that we know is just equal to 1.

01:56 So we get 6 choose 1 times negative 2 to the negative 3rd to the first power.

02:05 For the third term, j equals 2.

02:07 So we get 6 choose 2, negative 2 to the negative 3rd.

02:12 To the second and hopefully by now you're seeing a pattern.

02:20 Now to find 620 all the way up to 626 we can use the 6th row of pascal's triangle.

02:32 So for the first term 6 to 0 equals 1 and negative 2 to the negative 3rd to the 0 power also equals 1.

02:41 So this first term is just 1...

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