The sum of the last 10 digits of the sum of the expression:
`(1^1 xx 2^2 xx 3^3 xx 4^4 xx 5^5) + (1^6 xx 2^7 xx 3^8 xx 4^9 xx 5^10) + (1^11 xx 2^12 xx 3^13 xx 4^14 xx 5^15) + ……`
`+(1^96 xx 2^97 xx 3^(98) xx 4^(99) xx 5^(100))` is :

Options

A. 16
B. 18
C. 20
D. none of these

Answer

To solve the given problem, we need to evaluate the expression and find the last 10 digits of the sum of the terms in the series. The expression is:

\[ (1^1 \times 2^2 \times 3^3 \times 4^4 \times 5^5) + (1^6 \times 2^7 \times 3^8 \times 4^9 \times 5^{10}) + (1^{11} \times 2^{12} \times 3^{13} \times 4^{14} \times 5^{15}) + \ldots + (1^{96} \times 2^{97} \times 3^{98} \times 4^{99} \times 5^{100}) \]

### Step 1: Identify the pattern in the terms
Each term in the series can be expressed as:
\[ 1^{n} \times 2^{n+1} \times 3^{n+2} \times 4^{n+3} \times 5^{n+4} \]
where \( n \) takes values from 1 to 96 in increments of 5.

### Step 2: Calculate the powers of 2 and 5
To find the last 10 digits, we need to consider how many trailing zeros are produced by the factors of 2 and 5 in each term. The number of trailing zeros in a product is determined by the minimum of the number of 2s and 5s in the factorization.

For each term:
- The power of 5 in the \( n \)-th term is \( n + 4 \).
- The power of 2 in the \( n \)-th term is \( n + 1 \).

### Step 3: Determine the maximum power of 5
The maximum power of 5 across all terms is:
- For \( n = 96 \): \( 5^{100} \) contributes 100 to the power of 5.

### Step 4: Determine the maximum power of 2
The maximum power of 2 across all terms is:
- For \( n = 96 \): \( 2^{97} \) contributes 97 to the power of 2.

### Step 5: Calculate the number of trailing zeros
Since the number of trailing zeros is determined by the minimum of the powers of 2 and 5:
\[ \text{Trailing Zeros} = \min(100, 97) = 97 \]
This means that the last 97 digits of the sum will be zeros.

### Step 6: Calculate the last 10 non-zero digits
To find the last 10 non-zero digits, we can consider the contributions from the first few terms:
1. Calculate \( 1^1 \times 2^2 \times 3^3 \times 4^4 \times 5^5 \):
- \( = 1 \times 4 \times 27 \times 625 = 67500 \)
2. Calculate \( 1^6 \times 2^7 \times 3^8 \times 4^9 \times 5^{10} \):
- \( = 1 \times 128 \times 6561 \times 262144 \times 9765625 \)

### Step 7: Sum the last 10 digits
Since we have determined that the last 97 digits are zeros, we only need to sum the last 10 digits of the non-zero contributions from the terms we calculated.

### Final Calculation
The sum of the last 10 digits from the contributions of the terms is:
- From \( 1^1 \times 2^2 \times 3^3 \times 4^4 \times 5^5 \): 67500 contributes 7 + 6 + 5 = 18.
- The other terms will contribute zeros due to trailing zeros.

Thus, the final answer for the sum of the last 10 digits is:
\[ \text{Sum} = 18 \]