11. Prove that the product of $$ n $$ geometric means between $$ a $$ and $$ b $$ to the nth power of the single geometric mean between them. For what value of $$ n, \frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}} $$ is the geometric mean betweo iw?

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### **Problem Statement**

1. **Proof**: Show that the product of $$ n $$ geometric means between $$ a $$ and $$ b $$ is equal to the $$ n $$th power of the single geometric mean between them.
2. **Find $$ n $$**: Determine the value of $$ n $$ for which 
   $$
   \frac{a^{n+1} + b^{n+1}}{a^n + b^n}
   $$
   equals the geometric mean between $$ a $$ and $$ b $$.

---

### **1. Proving the Product of $$ n $$ Geometric Means**

**Understanding Geometric Means:**

Suppose there are $$ n $$ geometric means between $$ a $$ and $$ b $$. This sequence forms a geometric progression:
$$
a, G_1, G_2, \dots, G_n, b
$$
Here, $$ G_1, G_2, \dots, G_n $$ are the geometric means, and the common ratio is $$ r $$.

**Expressing $$ b $$ in Terms of $$ a $$ and $$ r $$:**
$$
b = a \cdot r^{n+1}
$$
$$
\Rightarrow r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}
$$

**Product of the $$ n $$ Geometric Means:**
$$
G_1 \times G_2 \times \dots \times G_n = a^n \cdot r^{1 + 2 + \dots + n} = a^n \cdot r^{\frac{n(n+1)}{2}}
$$
Substituting $$ r $$:
$$
= a^n \cdot \left(\frac{b}{a}\right)^{\frac{n}{2}} = (a^{\frac{n}{2}} \cdot b^{\frac{n}{2}}) = (ab)^{\frac{n}{2}}
$$

**Single Geometric Mean:**
$$
\sqrt{ab}
$$
**Raising to the $$ n $$th Power:**
$$
(\sqrt{ab})^n = (ab)^{\frac{n}{2}}
$$

**Conclusion:**
$$
G_1 \times G_2 \times \dots \times G_n = (\sqrt{ab})^n
$$
Thus, the product of $$ n $$ geometric means between $$ a $$ and $$ b $$ is indeed equal to the $$ n $$th power of the single geometric mean between them.

---

### **2. Finding the Value of $$ n $$**

We need to find $$ n $$ such that:
$$
\frac{a^{n+1} + b^{n+1}}{a^n + b^n} = \sqrt{ab}
$$

**Let’s Introduce a Substitution:**
Let $$ k = \frac{a}{b} $$. Assume $$ b \neq 0 $$ and $$ k \neq 1 $$ (since if $$ k = 1 $$, $$ a = b $$, and the equation trivially holds for any $$ n $$).

**Rewrite the Equation in Terms of $$ k $$:**
$$
\frac{a^{n+1} + b^{n+1}}{a^n + b^n} = b \cdot \frac{k^{n+1} + 1}{k^n + 1} = \sqrt{ab} = b \sqrt{k}
$$
Divide both sides by $$ b $$:
$$
\frac{k^{n+1} + 1}{k^n + 1} = \sqrt{k}
$$

**Cross-Multiplying:**
$$
k^{n+1} + 1 = \sqrt{k} \cdot (k^n + 1)
$$
$$
k^{n+1} + 1 = k^{n + \frac{1}{2}} + \sqrt{k}
$$
Rearranging:
$$
k^{n+1} - k^{n + \frac{1}{2}} + 1 - \sqrt{k} = 0
$$
Factor by grouping:
$$
k^{n + \frac{1}{2}}(k^{\frac{1}{2}} - 1) + (1 - k^{\frac{1}{2}}) = 0
$$
$$
(k^{\frac{1}{2}} - 1)(k^{n + \frac{1}{2}} - 1) = 0
$$
Given that $$ k \neq 1 $$, the only solution comes from:
$$
k^{n + \frac{1}{2}} = 1
$$
Since $$ k > 0 $$ and $$ k \neq 1 $$, the exponent must be zero:
$$
n + \frac{1}{2} = 0
$$
$$
n = -\frac{1}{2}
$$

**Conclusion:**
The value of $$ n $$ that satisfies the given equation is $$ n = -\frac{1}{2} $$.

---

### **Final Answer**

1. **Proof**: The product of $$ n $$ geometric means between $$ a $$ and $$ b $$ is equal to $$ (\sqrt{ab})^n $$.

2. **Value of $$ n $$**: The value of $$ n $$ for which 
   $$
   \frac{a^{n+1} + b^{n+1}}{a^n + b^n} 
   $$
   equals the geometric mean between $$ a $$ and $$ b $$ is $$ n = -\frac{1}{2} $$.

**Answer**:
After simplifying, the required value is –½. Thus,

1. The product of n geometric means between a and b is (√(ab))ⁿ.

2. When n = −½, (aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) is equal to the geometric mean of a and b.
The value of $$ n $$ is $$-\frac{1}{2}$$.

Prove that the product of geometric means between and
to the nth power of the single geometric mean between them.
For what value of is the geometric mean betweo
iw?

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Prove that the product of geometric means between and to the nth power of the single geometric mean between them. For what value of is the geometric mean betweo iw?