https://file.aiquickdraw.com/mathbot/file/1725313048655hlm57whc.jpg

To solve the problem, let's break it down step by step.

### Problem Statement:
We are given the binomial expansion of \((1 + x)^{15}\). The coefficients of the 5th, 6th, and 7th terms in this expansion, when expressed in ascending powers of \(x\), are in linear sequence. We need to find the possible values of \(x\).

### Step 1: Identify the General Term in the Binomial Expansion
The general term in the binomial expansion of \((1 + x)^{15}\) is given by:
\[
T_{r+1} = \binom{15}{r} x^r
\]
where \( \binom{15}{r} \) is the binomial coefficient.

### Step 2: Express the Required Terms
- The 5th term corresponds to \( r = 4 \):
  \[
  T_5 = \binom{15}{4} x^4
  \]
- The 6th term corresponds to \( r = 5 \):
  \[
  T_6 = \binom{15}{5} x^5
  \]
- The 7th term corresponds to \( r = 6 \):
  \[
  T_7 = \binom{15}{6} x^6
  \]

### Step 3: Given Condition
We are told that the coefficients of these terms are in linear sequence. Therefore:
\[
\text{Coefficient of } T_6 - \text{Coefficient of } T_5 = \text{Coefficient of } T_7 - \text{Coefficient of } T_6
\]
Substituting the coefficients:
\[
\binom{15}{5} x^5 - \binom{15}{4} x^4 = \binom{15}{6} x^6 - \binom{15}{5} x^5
\]

### Step 4: Simplify the Equation
Divide the entire equation by \( x^4 \) (assuming \( x \neq 0 \)):
\[
\binom{15}{5} x - \binom{15}{4} = \binom{15}{6} x^2 - \binom{15}{5} x
\]
Rearrange terms:
\[
\binom{15}{6} x^2 - 2\binom{15}{5} x + \binom{15}{4} = 0
\]
This is a quadratic equation in \( x \).

### Step 5: Substitute the Binomial Coefficients
Now, substitute the binomial coefficients:
\[
\binom{15}{4} = 1365, \quad \binom{15}{5} = 3003, \quad \binom{15}{6} = 5005
\]
Substitute into the quadratic equation:
\[
5005x^2 - 2(3003)x + 1365 = 0
\]
Simplify further:
\[
5005x^2 - 6006x + 1365 = 0
\]

### Step 6: Solve the Quadratic Equation
Divide the entire equation by 5 for simplification:
\[
1001x^2 - 1201.2x + 273 = 0
\]
Now, solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1001 \), \( b = -1201.2 \), and \( c = 273 \).

### Step 7: Calculate the Roots
Using the values \( a \), \( b \), and \( c \), compute the discriminant and solve for \( x \).

Let me calculate the roots now.The possible values of \( x \) are approximately \( x = 0.895 \) and \( x = 0.305 \).

These are the roots of the quadratic equation derived from the condition that the coefficients of the 5th, 6th, and 7th terms are in linear sequence.

Would you like more details on this solution or have any other questions? Here are some related questions:

1. How would the solution change if the binomial was \((1 - x)^{15}\)?
2. Can you explain the significance of the binomial coefficient in expansions?
3. How do you determine the general term in a binomial expansion?
4. What happens if the coefficients are in a geometric sequence instead?
5. Can you apply this method to solve for another value of \( n \) in \((1 + x)^n\)?

**Tip:** Always double-check your quadratic formula calculations, especially when dealing with non-integer coefficients, to avoid rounding errors.

Explore how to find the values of x in the binomial expansion (1 + x)^15 where the coefficients of the 5th, 6th, and 7th terms are in linear sequence. Detailed step-by-step solution involving the binomial theorem and quadratic equations. Suitable for advanced high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation