Using The Binomial Theorem, Obtain The Expansion Of $(1+3x)^6 + (1-3x)^6$.Use The Above Result To Calculate The Value Of $(1.03)^6 + (0.97)^6$, Correct To 5 Decimal Places.

Introduction

The binomial theorem is a powerful tool in mathematics that allows us to expand expressions of the form $(a + b)^n$, where $a$ and $b$ are constants and $n$ is a positive integer. In this article, we will use the binomial theorem to expand the expressions $(1+3x)^6$ and $(1-3x)^6$, and then use the resulting expansions to calculate the value of $(1.03)^6 + (0.97)^6$, correct to 5 decimal places.

The Binomial Theorem

The binomial theorem states that for any positive integer $n$, the expansion of $(a + b)^n$ is given by:

$$(a + b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \cdots + \binom{n}{n-1} a^1 b^{n-1} + \binom{n}{n} a^0 b^n$$

where $\binom{n}{k}$ is the binomial coefficient, defined as:

$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$

Expanding $(1+3x)^6$

Using the binomial theorem, we can expand $(1+3x)^6$ as follows:

$$(1+3x)^6 = \binom{6}{0} 1^6 (3x)^0 + \binom{6}{1} 1^5 (3x)^1 + \binom{6}{2} 1^4 (3x)^2 + \binom{6}{3} 1^3 (3x)^3 + \binom{6}{4} 1^2 (3x)^4 + \binom{6}{5} 1^1 (3x)^5 + \binom{6}{6} 1^0 (3x)^6$$

Simplifying the expression, we get:

$$(1+3x)^6 = 1 + 6(3x) + 15(3x)^2 + 20(3x)^3 + 15(3x)^4 + 6(3x)^5 + (3x)^6$$

$$(1+3x)^6 = 1 + 18x + 45x^2 + 60x^3 + 45x^4 + 18x^5 + 3x^6$$

Expanding $(1-3x)^6$

Using the binomial theorem, we can expand $(1-3x)^6$ as follows:

$$(1-3x)^6 = \binom{6}{0} 1^6 (-3x)^0 + \binom{6}{1} 1^5 (-3x)^1 + \binom{6}{2} 1^4 (-3x)^2 + \binom{6}{3} 1^3 (-3x)^3 + \binom{6}{4} 1^2 (-3x)^4 + \binom{6}{5} 1^1 (-3x)^5 + \binom{6}{6} 1^0 (-3x)^6$$

Simplifying the expression, we get:

$$(1-3x)^6 = 1 - 6(3x) + 15(3x)^2 - 20(3x)^3 + 15(3x)^4 - 6(3x)^5 + (3x)^6$$

$$(1-3x)^6 = 1 - 18x + 45x^2 - 60x^3 + 45x^4 - 18x^5 + 3x^6$$

Adding the Expansions

Now that we have expanded both $(1+3x)^6$ and $(1-3x)^6$, we can add the two expressions together:

$$(1+3x)^6 + (1-3x)^6 = (1 + 18x + 45x^2 + 60x^3 + 45x^4 + 18x^5 + 3x^6) + (1 - 18x + 45x^2 - 60x^3 + 45x^4 - 18x^5 + 3x^6)$$

Simplifying the expression, we get:

$$(1+3x)^6 + (1-3x)^6 = 2 + 90x^2 + 3x^6$$

Calculating the Value of $(1.03)^6 + (0.97)^6$

Now that we have the expansion of $(1+3x)^6 + (1-3x)^6$, we can use it to calculate the value of $(1.03)^6 + (0.97)^6$. To do this, we need to substitute $x = 0.01$ into the expansion:

$$(1+3x)^6 + (1-3x)^6 = 2 + 90x^2 + 3x^6$$

Substituting $x = 0.01$, we get:

$$(1.03)^6 + (0.97)^6 = 2 + 90(0.01)^2 + 3(0.01)^6$$

Simplifying the expression, we get:

$$(1.03)^6 + (0.97)^6 = 2 + 0.009 + 0.0000003$$

$$(1.03)^6 + (0.97)^6 = 2.0090003$$

Rounding to 5 decimal places, we get:

$$(1.03)^6 + (0.97)^6 = 2.00900$$

Conclusion

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Q: What is the binomial theorem?

A: The binomial theorem is a powerful tool in mathematics that allows us to expand expressions of the form $(a + b)^n$, where $a$ and $b$ are constants and $n$ is a positive integer.

Q: How do I use the binomial theorem to expand expressions?

A: To use the binomial theorem to expand expressions, you need to follow these steps:

1. Identify the values of $a$, $b$, and $n$ in the expression.
2. Use the binomial theorem formula to expand the expression.
3. Simplify the resulting expression.

Q: What is the binomial coefficient?

A: The binomial coefficient is a number that appears in the binomial theorem formula. It is defined as:

$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$

Q: How do I calculate the binomial coefficient?

A: To calculate the binomial coefficient, you need to follow these steps:

1. Calculate the factorial of $n$.
2. Calculate the factorial of $k$.
3. Calculate the factorial of $n-k$.
4. Divide the result from step 1 by the product of the results from steps 2 and 3.

Q: What is the expansion of $(1+3x)^6$?

A: The expansion of $(1+3x)^6$ is given by:

$$(1+3x)^6 = 1 + 6(3x) + 15(3x)^2 + 20(3x)^3 + 15(3x)^4 + 6(3x)^5 + (3x)^6$$

$$(1+3x)^6 = 1 + 18x + 45x^2 + 60x^3 + 45x^4 + 18x^5 + 3x^6$$

Q: What is the expansion of $(1-3x)^6$?

A: The expansion of $(1-3x)^6$ is given by:

$$(1-3x)^6 = 1 - 6(3x) + 15(3x)^2 - 20(3x)^3 + 15(3x)^4 - 6(3x)^5 + (3x)^6$$

$$(1-3x)^6 = 1 - 18x + 45x^2 - 60x^3 + 45x^4 - 18x^5 + 3x^6$$

Q: How do I add the expansions of $(1+3x)^6$ and $(1-3x)^6$?

A: To add the expansions of $(1+3x)^6$ and $(1-3x)^6$, you need to follow these steps:

1. Combine like terms.
2. Simplify the resulting expression.

The expansion of $(1+3x)^6 + (1-3x)^6$ is given by:

$$(1+3x)^6 + (1-3x)^6 = 2 + 90x^2 + 3x^6$$

Q: How do I use the expansion of $(1+3x)^6 + (1-3x)^6$ to calculate the value of $(1.03)^6 + (0.97)^6$?

A: To use the expansion of $(1+3x)^6 + (1-3x)^6$ to calculate the value of $(1.03)^6 + (0.97)^6$, you need to follow these steps:

1. Substitute $x = 0.01$ into the expansion.
2. Simplify the resulting expression.

The value of $(1.03)^6 + (0.97)^6$ is approximately $2.00900$.

Q: What is the final answer to the problem?

A: The final answer to the problem is $2.00900$.