A) Find The First Three Terms, In Ascending Powers Of X X X , Of The Binomial Expansion Of ( 1 + P X ) 30 (1 + Px)^{30} ( 1 + P X ) 30 , Where P P P Is A Positive Constant.b) Given That In This Expansion The Coefficient Of X 2 X^2 X 2 Is 29 Times The

Introduction

The binomial expansion is a powerful tool in mathematics, used 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 explore the binomial expansion of the expression $(1 + px)^{30}$, where $p$ is a positive constant. We will find the first three terms of the expansion in ascending powers of $x$ and analyze the coefficient of $x^2$.

Binomial Expansion Formula

The binomial expansion formula is given by:

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

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

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

First Three Terms of the Expansion

To find the first three terms of the expansion of $(1 + px)^{30}$, we will use the binomial expansion formula. We have:

$$(1 + px)^{30} = \binom{30}{0} 1^{30} (px)^0 + \binom{30}{1} 1^{29} (px)^1 + \binom{30}{2} 1^{28} (px)^2 + \cdots$$

Using the binomial coefficient formula, we can simplify the expression:

$$(1 + px)^{30} = 1 + 30p(x) + \frac{30 \cdot 29}{2} p^2(x^2) + \cdots$$

Therefore, the first three terms of the expansion are:

$$1 + 30px + 435p^2x^2$$

Coefficient of $x^2$

The coefficient of $x^2$ in the expansion is $435p^2$. We are given that this coefficient is 29 times the coefficient of $x^2$ in the expansion of $(1 + px)^{29}$. To find the coefficient of $x^2$ in the expansion of $(1 + px)^{29}$, we will use the binomial expansion formula:

$$(1 + px)^{29} = \binom{29}{0} 1^{29} (px)^0 + \binom{29}{1} 1^{28} (px)^1 + \binom{29}{2} 1^{27} (px)^2 + \cdots$$

Using the binomial coefficient formula, we can simplify the expression:

$$(1 + px)^{29} = 1 + 29p(x) + \frac{29 \cdot 28}{2} p^2(x^2) + \cdots$$

Therefore, the coefficient of $x^2$ in the expansion of $(1 + px)^{29}$ is $\frac{29 \cdot 28}{2} p^2 = 406p^2$.

Equating Coefficients

We are given that the coefficient of $x^2$ in the expansion of $(1 + px)^{30}$ is 29 times the coefficient of $x^2$ in the expansion of $(1 + px)^{29}$. Therefore, we can set up the equation:

$$435p^2 = 29 \cdot 406p^2$$

Simplifying the equation, we get:

$$435 = 29 \cdot 406$$

This equation is not true, which means that the given condition is not satisfied. Therefore, the expansion of $(1 + px)^{30}$ does not have a coefficient of $x^2$ that is 29 times the coefficient of $x^2$ in the expansion of $(1 + px)^{29}$.

Conclusion

In this article, we have explored the binomial expansion of the expression $(1 + px)^{30}$, where $p$ is a positive constant. We have found the first three terms of the expansion in ascending powers of $x$ and analyzed the coefficient of $x^2$. We have also shown that the given condition is not satisfied, which means that the expansion of $(1 + px)^{30}$ does not have a coefficient of $x^2$ that is 29 times the coefficient of $x^2$ in the expansion of $(1 + px)^{29}$.

References

• Binomial expansion formula: $(a + b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \cdots + \binom{n}{n-1} a^1 b^{n-1} + \binom{n}{n} a^0 b^n$
• Binomial coefficient formula: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$
  Binomial Expansion and Coefficient Analysis: Q&A =====================================================

Introduction

In our previous article, we explored the binomial expansion of the expression $(1 + px)^{30}$, where $p$ is a positive constant. We found the first three terms of the expansion in ascending powers of $x$ and analyzed the coefficient of $x^2$. In this article, we will answer some frequently asked questions related to binomial expansion and coefficient analysis.

Q: What is the binomial expansion formula?

A: The binomial expansion formula is given by:

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

Q: What is the binomial coefficient formula?

A: The binomial coefficient formula is given by:

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

Q: How do I find the first three terms of the binomial expansion of $(1 + px)^{30}$?

A: To find the first three terms of the binomial expansion of $(1 + px)^{30}$, we will use the binomial expansion formula. We have:

$$(1 + px)^{30} = \binom{30}{0} 1^{30} (px)^0 + \binom{30}{1} 1^{29} (px)^1 + \binom{30}{2} 1^{28} (px)^2 + \cdots$$

Using the binomial coefficient formula, we can simplify the expression:

$$(1 + px)^{30} = 1 + 30p(x) + \frac{30 \cdot 29}{2} p^2(x^2) + \cdots$$

Therefore, the first three terms of the expansion are:

$$1 + 30px + 435p^2x^2$$

Q: What is the coefficient of $x^2$ in the expansion of $(1 + px)^{30}$?

A: The coefficient of $x^2$ in the expansion of $(1 + px)^{30}$ is $435p^2$.

Q: How do I find the coefficient of $x^2$ in the expansion of $(1 + px)^{29}$?

A: To find the coefficient of $x^2$ in the expansion of $(1 + px)^{29}$, we will use the binomial expansion formula:

$$(1 + px)^{29} = \binom{29}{0} 1^{29} (px)^0 + \binom{29}{1} 1^{28} (px)^1 + \binom{29}{2} 1^{27} (px)^2 + \cdots$$

Using the binomial coefficient formula, we can simplify the expression:

$$(1 + px)^{29} = 1 + 29p(x) + \frac{29 \cdot 28}{2} p^2(x^2) + \cdots$$

Therefore, the coefficient of $x^2$ in the expansion of $(1 + px)^{29}$ is $\frac{29 \cdot 28}{2} p^2 = 406p^2$.

Q: What is the relationship between the coefficients of $x^2$ in the expansions of $(1 + px)^{30}$ and $(1 + px)^{29}$?

A: We are given that the coefficient of $x^2$ in the expansion of $(1 + px)^{30}$ is 29 times the coefficient of $x^2$ in the expansion of $(1 + px)^{29}$. Therefore, we can set up the equation:

$$435p^2 = 29 \cdot 406p^2$$

Simplifying the equation, we get:

$$435 = 29 \cdot 406$$

This equation is not true, which means that the given condition is not satisfied. Therefore, the expansion of $(1 + px)^{30}$ does not have a coefficient of $x^2$ that is 29 times the coefficient of $x^2$ in the expansion of $(1 + px)^{29}$.

Conclusion

In this article, we have answered some frequently asked questions related to binomial expansion and coefficient analysis. We have also shown that the given condition is not satisfied, which means that the expansion of $(1 + px)^{30}$ does not have a coefficient of $x^2$ that is 29 times the coefficient of $x^2$ in the expansion of $(1 + px)^{29}$.

References

• Binomial expansion formula: $(a + b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \cdots + \binom{n}{n-1} a^1 b^{n-1} + \binom{n}{n} a^0 b^n$
• Binomial coefficient formula: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$