Question

Use the Binomial Theorem to find the numerical value of \((1.001)^{5}\) correct to five decimal places. \(\left[\right.\) Hint: \(\left.(1.001)^{5}=\left(1+10^{-3}\right)^{5}\right]\)

Short answer

1.00501

Step-by-step solution

Identify the binomial expression

Recognize that \( (1.001)^{5} \) can be rewritten using the hint as \( (1 + 10^{-3})^{5} \). This is our binomial expression.

Apply the Binomial Theorem

The Binomial Theorem states that \[ (a + b)^{n} = \binom{n}{0} a^{n}b^{0} + \binom{n}{1} a^{n-1}b^{1} + \binom{n}{2} a^{n-2}b^{2} + \binom{n}{3} a^{n-3}b^{3} + \binom{n}{4} a^{n-4}b^{4} + \binom{n}{5} a^{n-5}b^{5} \].

Substitute the values into the Binomial Theorem

Substitute \( a = 1 \) , \( b = 10^{-3} \), and \( n = 5 \):\[ (1 + 10^{-3})^5 = \binom{5}{0} (1)^5 (10^{-3})^0 + \binom{5}{1} (1)^4 (10^{-3})^1 + \binom{5}{2} (1)^3 (10^{-3})^2 + \binom{5}{3} (1)^2 (10^{-3})^3 + \binom{5}{4} (1)^1 (10^{-3})^4 + \binom{5}{5} (1)^0 (10^{-3})^5 \]

Calculate each term

Calculate each term separately: \ \[ \binom{5}{0} (1)^5 (10^{-3})^0 = 1 \] \[ \binom{5}{1} (1)^4 (10^{-3})^1 = 5 \times 10^{-3} = 0.005 \] \[ \binom{5}{2} (1)^3 (10^{-3})^2 = 10 \times 10^{-6} = 0.00001 \] \[ \binom{5}{3} (1)^2 (10^{-3})^3 = 10 \times 10^{-9} = 0.00000001 \] \[ \binom{5}{4} (1)^1 (10^{-3})^4 = 5 \times 10^{-12} = 0.000000000005 \] \[ \binom{5}{5} (1)^0 (10^{-3})^5 = 1 \times 10^{-15} = 0.000000000000001 \]

Sum the terms

Add all terms together to find the numerical value: \[(1 + 10^{-3})^5 = 1 + 0.005 + 0.00001 + 0.00000001 + 0.000000000005 + 0.000000000000001 \approx 1.00501 \] when rounded to five decimal places.

Key concepts

Algebra

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating those symbols. In algebra, symbols (like x or y) are used to represent numbers or values in equations. This allows us to formulate expressions and solve equations.
Algebra is essential in many areas of mathematics, including calculus and geometry. It helps us solve real-world problems by modeling situations with equations and finding unknown values.
When working with more complex algebraic expressions and operations, understanding the basics of algebra is key. For example, the Binomial Theorem, which expands expressions like \( (a + b)^n \), is a crucial algebraic tool.

Polynomial Expansion

Polynomial expansion is the process of expressing a polynomial as a sum of terms. The Binomial Theorem is a powerful tool for polynomial expansion. It allows us to expand binomials raised to a power, such as \( (1 + 10^{-3})^5 \).
According to the Binomial Theorem, we can expand this as follows:
\[ (a + b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + ... + \binom{n}{n} a^0 b^n \]
Here, \ a \ and \ b \ are any numbers, and \ n \ is a positive integer. The symbol \ \binom{n}{k} \ represents a combination, which shows the number of ways to choose \ k \ items from \ n \ items.
In our exercise, we used \ a = 1 \ and \ b = 10^{-3} \ with \ n = 5 \ to expand \ (1 + 10^{-3})^5 \ and then calculated each term individually. By summing these terms, we approximated the value to five decimal places.

Combinations

Combinations are a fundamental concept in combinatorics, the branch of mathematics dealing with counting. A combination is a way to select items from a larger set where the order of selection does not matter.
The combination formula is given by:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Here, \ n \ is the total number of items, \ k \ is the number of items to choose, and \ n! \ (n factorial) is the product of all positive integers up to \ n \.
In our example, we used combinations to calculate each term in the binomial expansion. For instance, \ \binom{5}{2} \ is the number of ways to choose 2 items from 5, which is equal to:
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
Understanding combinations helps in calculating each term in polynomial expansion accurately. This is crucial, as even small errors in intermediate calculations can lead to incorrect final results.