Let a,b,c,d and e be positive real numbers such that `a+b+c+d+e=15` and `ab^2c^3d^4e^5=(120)^3xx50`. Then the value of `a^2+b^2+c^2+d^2+e^2` is ___________.

Explanation

To solve the problem, we need to find the value of \( a^2 + b^2 + c^2 + d^2 + e^2 \) given the conditions:

1. \( a + b + c + d + e = 15 \)
2. \( ab^2c^3d^4e^5 = (120)^3 \times 50 \)

### Step 1: Set up the equations
We know that:
- \( a + b + c + d + e = 15 \)
- \( ab^2c^3d^4e^5 = 120^3 \times 50 \)

### Step 2: Calculate \( 120^3 \times 50 \)
First, we simplify \( 120^3 \times 50 \):
\[ 120 = 2^3 \times 3 \times 5 \]
Thus,
\[ 120^3 = (2^3 \times 3 \times 5)^3 = 2^9 \times 3^3 \times 5^3 \]
Now, multiplying by \( 50 \):
\[ 50 = 2 \times 5^2 \]
So,
\[ 120^3 \times 50 = 2^9 \times 3^3 \times 5^3 \times 2 \times 5^2 = 2^{10} \times 3^3 \times 5^5 \]

### Step 3: Apply the AM-GM inequality
Using the AM-GM inequality:
\[ \frac{a + 2b + 3c + 4d + 5e}{1 + 2 + 3 + 4 + 5} \geq \sqrt[15]{ab^2c^3d^4e^5} \]
The left side simplifies to:
\[ \frac{a + 2b + 3c + 4d + 5e}{15} \]
The right side becomes:
\[ \sqrt[15]{120^3 \times 50} = \sqrt[15]{2^{10} \times 3^3 \times 5^5} \]

### Step 4: Find the equality condition
The equality in AM-GM holds when:
\[ a = \frac{1}{15} \cdot 1, \quad b = \frac{2}{15} \cdot 2, \quad c = \frac{3}{15} \cdot 3, \quad d = \frac{4}{15} \cdot 4, \quad e = \frac{5}{15} \cdot 5 \]
Let \( \lambda \) be a common factor:
\[ a = \lambda, \quad b = 2\lambda, \quad c = 3\lambda, \quad d = 4\lambda, \quad e = 5\lambda \]
Then:
\[ \lambda + 2\lambda + 3\lambda + 4\lambda + 5\lambda = 15\lambda = 15 \implies \lambda = 1 \]
Thus:
\[ a = 1, \quad b = 2, \quad c = 3, \quad d = 4, \quad e = 5 \]

### Step 5: Calculate \( a^2 + b^2 + c^2 + d^2 + e^2 \)
Now, we compute:
\[ a^2 + b^2 + c^2 + d^2 + e^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 \]
Calculating each term:
\[ 1^2 = 1, \quad 2^2 = 4, \quad 3^2 = 9, \quad 4^2 = 16, \quad 5^2 = 25 \]
Adding them together:
\[ 1 + 4 + 9 + 16 + 25 = 55 \]

### Final Answer
Thus, the value of \( a^2 + b^2 + c^2 + d^2 + e^2 \) is \( \boxed{55} \).