(x+1)^7, the third term in the expansion is the mean of the second term and the fourth term in the expansion. Find the possible values of x

We are given the expression \((x+1)^7\), and we are asked to find the possible values of \(x\) such that the third term in the binomial expansion is the mean of the second term and the fourth term.

### Step 1: Binomial Expansion Formula
The binomial expansion of \((x + 1)^7\) follows the general binomial expansion formula:

\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]

In our case, \(a = x\), \(b = 1\), and \(n = 7\). The general term (the \(k\)-th term) in the expansion is given by:

\[
T_{k+1} = \binom{7}{k} x^{7-k} \cdot 1^k = \binom{7}{k} x^{7-k}
\]

### Step 2: Identify the Second, Third, and Fourth Terms

- **Second term (\(T_2\))**: This corresponds to \(k = 1\),
  \[
  T_2 = \binom{7}{1} x^6 = 7x^6
  \]

- **Third term (\(T_3\))**: This corresponds to \(k = 2\),
  \[
  T_3 = \binom{7}{2} x^5 = \frac{7 \times 6}{2} x^5 = 21x^5
  \]

- **Fourth term (\(T_4\))**: This corresponds to \(k = 3\),
  \[
  T_4 = \binom{7}{3} x^4 = \frac{7 \times 6 \times 5}{6} x^4 = 35x^4
  \]

### Step 3: Set up the Equation for the Mean
We are told that the third term is the mean of the second and fourth terms. This means:

\[
T_3 = \frac{T_2 + T_4}{2}
\]

Substitute the values of \(T_2\), \(T_3\), and \(T_4\):

\[
21x^5 = \frac{7x^6 + 35x^4}{2}
\]

### Step 4: Solve for \(x\)

Multiply both sides by 2 to eliminate the fraction:

\[
42x^5 = 7x^6 + 35x^4
\]

Rearrange the equation:

\[
7x^6 - 42x^5 + 35x^4 = 0
\]

Factor out the common term \(7x^4\):

\[
7x^4(x^2 - 6x + 5) = 0
\]

Now solve the factored equation:

1. \(7x^4 = 0\) gives \(x = 0\).
2. Solve \(x^2 - 6x + 5 = 0\) using the quadratic formula:
   \[
   x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(5)}}{2(1)} = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2}
   \]
   So, \(x = \frac{6 + 4}{2} = 5\) or \(x = \frac{6 - 4}{2} = 1\).

### Step 5: Conclusion

The possible values of \(x\) are:

\[
x = 0, 1, 5
\]

Would you like further details on any part of the solution? 

Here are 5 related questions:
1. How is the binomial expansion formula derived?
2. What happens if we expand \((x + 1)^n\) for different values of \(n\)?
3. How would the solution change if the powers in the equation were different?
4. Can we use the same method for other binomial powers (e.g., \((x+2)^5\))?
5. What is the importance of solving such types of equations in calculus?

**Tip:** Always check the highest-degree terms when factoring polynomials to simplify your calculations.

(x+1)^7, the third term in the expansion is the mean of the second term and the fourth term in the expansion. Find the possible values of x.

Solve the binomial expansion problem where the third term in (x+1)^7 is the mean of the second and fourth terms. This problem covers binomial expansion, polynomial factorization, and quadratic equations, suitable for students in Grades 10-12.

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