Determine The First Three Terms In The Binomial Expansion Of $\left(1 - \frac{1}{2^x}\right)^{\frac{1}{2}}$.i) By Substituting $x = \frac{1}{50}$ Into The Result, Determine The Value Of $\sqrt{11}$ Correct To Four Decimal Places.

Introduction

The binomial expansion is a powerful tool in mathematics that allows us to expand expressions of the form $(a + b)^n$. In this article, we will determine the first three terms in the binomial expansion of $\left(1 - \frac{1}{2^x}\right)^{\frac{1}{2}}$. We will then use this result to find the value of $\sqrt{11}$ correct to four decimal places by substituting $x = \frac{1}{50}$ into the result.

Binomial Expansion Formula

The binomial expansion formula is given by:

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

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

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

Determine the First Three Terms

To determine the first three terms in the binomial expansion of $\left(1 - \frac{1}{2^x}\right)^{\frac{1}{2}}$, we can use the binomial expansion formula with $a = 1$, $b = -\frac{1}{2^x}$, and $n = \frac{1}{2}$.

Using the binomial expansion formula, we get:

$$\left(1 - \frac{1}{2^x}\right)^{\frac{1}{2}} = \binom{\frac{1}{2}}{0} 1^{\frac{1}{2}} \left(-\frac{1}{2^x}\right)^0 + \binom{\frac{1}{2}}{1} 1^{\frac{1}{2}-1} \left(-\frac{1}{2^x}\right)^1 + \binom{\frac{1}{2}}{2} 1^{\frac{1}{2}-2} \left(-\frac{1}{2^x}\right)^2$$

Simplifying, we get:

$$\left(1 - \frac{1}{2^x}\right)^{\frac{1}{2}} = 1 + \frac{1}{2} \left(-\frac{1}{2^x}\right) + \frac{\frac{1}{2} \left(\frac{1}{2}\right)}{2!} \left(-\frac{1}{2^x}\right)^2$$

$$= 1 - \frac{1}{2 \cdot 2^x} + \frac{1}{2 \cdot 2^{2x}}$$

Substitute $x = \frac{1}{50}$

Now, we will substitute $x = \frac{1}{50}$ into the result to find the value of $\sqrt{11}$ correct to four decimal places.

Substituting $x = \frac{1}{50}$, we get:

$$\left(1 - \frac{1}{2^{\frac{1}{50}}}\right)^{\frac{1}{2}} = 1 - \frac{1}{2 \cdot 2^{\frac{1}{50}}} + \frac{1}{2 \cdot 2^{2 \cdot \frac{1}{50}}}$$

$$= 1 - \frac{1}{2 \cdot 2^{\frac{1}{50}}} + \frac{1}{2 \cdot 2^{\frac{1}{25}}}$$

Calculate the Value of $\sqrt{11}$

To calculate the value of $\sqrt{11}$, we need to find the value of $\left(1 - \frac{1}{2^{\frac{1}{50}}}\right)^{\frac{1}{2}}$ correct to four decimal places.

Using a calculator, we get:

$$\left(1 - \frac{1}{2^{\frac{1}{50}}}\right)^{\frac{1}{2}} \approx 1.048808$$

Now, we can use this result to find the value of $\sqrt{11}$ correct to four decimal places.

We know that:

$$\left(1 - \frac{1}{2^{\frac{1}{50}}}\right)^{\frac{1}{2}} = \frac{\sqrt{11}}{\sqrt{10}}$$

Therefore, we can write:

$$\frac{\sqrt{11}}{\sqrt{10}} \approx 1.048808$$

Solving for $\sqrt{11}$, we get:

$$\sqrt{11} \approx 1.048808 \cdot \sqrt{10}$$

$$\approx 1.048808 \cdot 3.16227766$$

$$\approx 3.320061$$

Therefore, the value of $\sqrt{11}$ correct to four decimal places is approximately 3.3201.

Conclusion

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Q: What is the binomial expansion formula?

A: The binomial expansion formula is given by:

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

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

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

Q: How do you determine the first three terms in the binomial expansion of $\left(1 - \frac{1}{2^x}\right)^{\frac{1}{2}}$?

A: To determine the first three terms in the binomial expansion of $\left(1 - \frac{1}{2^x}\right)^{\frac{1}{2}}$, we can use the binomial expansion formula with $a = 1$, $b = -\frac{1}{2^x}$, and $n = \frac{1}{2}$.

Using the binomial expansion formula, we get:

$$\left(1 - \frac{1}{2^x}\right)^{\frac{1}{2}} = \binom{\frac{1}{2}}{0} 1^{\frac{1}{2}} \left(-\frac{1}{2^x}\right)^0 + \binom{\frac{1}{2}}{1} 1^{\frac{1}{2}-1} \left(-\frac{1}{2^x}\right)^1 + \binom{\frac{1}{2}}{2} 1^{\frac{1}{2}-2} \left(-\frac{1}{2^x}\right)^2$$

Simplifying, we get:

$$\left(1 - \frac{1}{2^x}\right)^{\frac{1}{2}} = 1 + \frac{1}{2} \left(-\frac{1}{2^x}\right) + \frac{1}{2 \cdot 2^{2x}}$$

Q: How do you substitute $x = \frac{1}{50}$ into the result to find the value of $\sqrt{11}$ correct to four decimal places?

A: To substitute $x = \frac{1}{50}$ into the result, we simply replace $x$ with $\frac{1}{50}$ in the expression:

$$\left(1 - \frac{1}{2^x}\right)^{\frac{1}{2}} = 1 - \frac{1}{2 \cdot 2^x} + \frac{1}{2 \cdot 2^{2x}}$$

Substituting $x = \frac{1}{50}$, we get:

$$\left(1 - \frac{1}{2^{\frac{1}{50}}}\right)^{\frac{1}{2}} = 1 - \frac{1}{2 \cdot 2^{\frac{1}{50}}} + \frac{1}{2 \cdot 2^{2 \cdot \frac{1}{50}}}$$

Q: How do you calculate the value of $\sqrt{11}$ correct to four decimal places?

A: To calculate the value of $\sqrt{11}$ correct to four decimal places, we can use the result from the previous question:

$$\left(1 - \frac{1}{2^{\frac{1}{50}}}\right)^{\frac{1}{2}} = \frac{\sqrt{11}}{\sqrt{10}}$$

Therefore, we can write:

$$\frac{\sqrt{11}}{\sqrt{10}} \approx 1.048808$$

Solving for $\sqrt{11}$, we get:

$$\sqrt{11} \approx 1.048808 \cdot \sqrt{10}$$

$$\approx 1.048808 \cdot 3.16227766$$

$$\approx 3.320061$$

Therefore, the value of $\sqrt{11}$ correct to four decimal places is approximately 3.3201.

Q: What is the significance of the binomial expansion in mathematics?

A: The binomial expansion is a powerful tool in mathematics that allows us to expand expressions of the form $(a + b)^n$. It has numerous applications in various fields, including algebra, geometry, and calculus. The binomial expansion is used to solve problems involving combinations, permutations, and probability.

Q: Can you provide more examples of binomial expansion?

A: Yes, here are a few more examples of binomial expansion:

• $(a + b)^2 = a^2 + 2ab + b^2$
• $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
• $(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$

These are just a few examples of the many possible binomial expansions. The binomial expansion can be used to solve a wide range of problems involving combinations and permutations.

Conclusion

In this Q&A article, we have discussed the binomial expansion and its applications in mathematics. We have also provided examples of binomial expansion and explained how to use it to solve problems involving combinations and permutations. The binomial expansion is a powerful tool that has numerous applications in various fields, and it is an essential concept in mathematics.