Problem 2
Question
Find \((2 x+3)^{5}\).
Step-by-Step Solution
Verified Answer
(2x+3)^5 = 32x^5 + 240x^4 + 720x^3 + 1080x^2 + 810x + 243
1Step 1 - Understand the Binomial Theorem
The Binomial Theorem states that \ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}. Here, we have \(a = 2x\), \(b = 3\), and \(n = 5\).
2Step 2 - Compute the Binomial Coefficients
Identify the binomial coefficients \(\binom{5}{k}\) for \ k = 0, 1, 2, 3, 4, 5. \ The coefficients are calculated as follows: \ \binom{5}{0} = 1\, \binom{5}{1} = 5\, \binom{5}{2} = 10\, \binom{5}{3} = 10\, \binom{5}{4} = 5\, \binom{5}{5} = 1.\
3Step 3 - Write Each Term Using the Binomial Theorem
Apply the binomial theorem formula: \ (2x+3)^5= \sum_{k=0}^{5} \binom{5}{k} (2x)^{5-k} (3)^{k}\. \ Calculate each term separately: \ \(k=0\): \binom{5}{0} (2x)^5 (3)^0 = (2x)^5 = 32x^5\, \(k=1\): \binom{5}{1} (2x)^4 (3)^1 = 5 \(16x^4\) \ (3) = 240x^4\, \(k=2\): \binom{5}{2} (2x)^3 (3)^2 = 10 \(8x^3\) \ (9) = 720x^3\, \(k=3\): \binom{5}{3} (2x)^2 (3)^3 = 10 \(4x^2\) \ (27) = 1080x^2\, \(k=4\): \binom{5}{4} (2x)^1 (3)^4 = 5 \(2x\) \ (81) = 810x\, \(k=5\): \binom{5}{5} (2x)^0 (3)^5 = (1) \ (243) = 243. \
4Step 4 - Combine All Terms
Sum up all the terms from Step 3: \[ \ (2x+3)^5 = 32x^5 + 240x^4 + 720x^3 + 1080x^2 + 810x + 243 \]
Key Concepts
Binomial CoefficientsPolynomial ExpansionMathematical Proofs
Binomial Coefficients
Binomial coefficients are essential components in the Binomial Theorem. They describe the coefficients that appear in the polynomial expansion of binomials.
The binomial coefficient \(\binom{n}{k}\) is also known as a combination or 'n choose k.' It represents the number of ways to choose k items from n items without regard to order.
The formula to calculate binomial coefficients is \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) where '!' denotes factorial, the product of all positive integers up to that number.
Let's see how this works in our example:
The binomial coefficient \(\binom{n}{k}\) is also known as a combination or 'n choose k.' It represents the number of ways to choose k items from n items without regard to order.
The formula to calculate binomial coefficients is \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) where '!' denotes factorial, the product of all positive integers up to that number.
Let's see how this works in our example:
- For \(\binom{5}{0} = \frac{5!}{0!(5-0)!} = 1\)
- For \(\binom{5}{1} = \frac{5!}{1!(5-1)!} = 5\)
- For \(\binom{5}{2} = \frac{5!}{2!(5-2)!} = 10\)
- For \(\binom{5}{3} = \frac{5!}{3!(5-3)!} = 10\)
- For \(\binom{5}{4} = \frac{5!}{4!(5-4)!} = 5\)
- For \(\binom{5}{5} = \frac{5!}{5!(5-5)!} = 1\)
Polynomial Expansion
The Binomial Theorem makes it straightforward to expand polynomials.
Essentially, the theorem tells us how to expand an expression of the form \( (a+b)^n \) into a sum involving terms of the form \( a \) and \ (b \) raised to various powers.
The general form of the binomial expansion is: \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}.
In the case of \( (2x+3)^5 \), using the binomial theorem, the expansion is:
\ (2x+3)^5 = \sum_{k=0}^{5} \binom{5}{k} (2x)^{5-k} (3)^{k} \.
Now, let’s break it down term by term:
All these terms added together give us the final expanded form: \ 32x^5 + 240x^4 + 720x^3 + 1080x^2 + 810x + 243 \.
Essentially, the theorem tells us how to expand an expression of the form \( (a+b)^n \) into a sum involving terms of the form \( a \) and \ (b \) raised to various powers.
The general form of the binomial expansion is: \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}.
In the case of \( (2x+3)^5 \), using the binomial theorem, the expansion is:
\ (2x+3)^5 = \sum_{k=0}^{5} \binom{5}{k} (2x)^{5-k} (3)^{k} \.
Now, let’s break it down term by term:
- For \ k=0 \:
\binom{5}{0} \( (2x)^5 (3)^0 = 32x^5 \) - For \ k=1 \:
\binom{5}{1} \( (2x)^4 (3)^1 = 240x^4 \) - For \ k=2 \:
\binom{5}{2} \( (2x)^3 (3)^2 = 720x^3 \) - For \ k=3 \:
\binom{5}{3} \( (2x)^2 (3)^3 = 1080x^2 \) - For \ k=4 \:
\binom{5}{4} \( (2x)^1 (3)^4 = 810x \) - For \ k=5 \:
\binom{5}{5} \( (2x)^0 (3)^5 = 243 \)
All these terms added together give us the final expanded form: \ 32x^5 + 240x^4 + 720x^3 + 1080x^2 + 810x + 243 \.
Mathematical Proofs
Understanding the principles behind mathematical proofs can clarify the steps we take.
In the context of the binomial theorem, proofs involve several concepts.
Another way is using combinatorial arguments, where we reinterpret the coefficients \( \binom{n}{k} \) as counting selections.
The deeper understanding of mathematical proofs brings clarity to solving more problems beyond binomials.
Developing a strong grasp of these concepts will greatly aid you in mathematical problem-solving.
In the context of the binomial theorem, proofs involve several concepts.
- First, the proof of the binomial theorem itself involves induction. We start by verifying that it holds for a basic case, such as n=1.
- Next, we assume it holds for a general case n=k. We then prove that if it holds for n=k, it must also hold for n=k+1.
Another way is using combinatorial arguments, where we reinterpret the coefficients \( \binom{n}{k} \) as counting selections.
The deeper understanding of mathematical proofs brings clarity to solving more problems beyond binomials.
Developing a strong grasp of these concepts will greatly aid you in mathematical problem-solving.
Other exercises in this chapter
Problem 1
Use the binomial theorem (with \(x=1000\) and \(y=1\) ) to calculate \(1001^{6} .\)
View solution Problem 3
If we select 1001 numbers from the set \(\\{1,2,3, \ldots, 2000\\}\) it is certain that there will be two numbers selected such that one divides the other. We c
View solution Problem 3
Provide an argument as to why an \(8 \times 8\) chessboard with two squares pruned from diagonally opposite corners cannot be tiled with dominoes.
View solution Problem 4
Given any set of 53 integers, show that there are two of them having the property that either their sum or their difference is evenly divisible by 103.
View solution