What is the value of `54^(3)+60^(3)+66^(3)+72^(3)+78^(3)+...+120^(3)?`

Options

A. 9142646
B. 9254646
C. 9245664
D. 9542664

Answer

To solve the problem of finding the value of \(54^3 + 60^3 + 66^3 + 72^3 + 78^3 + \ldots + 120^3\), we can follow these steps:

### Step 1: Identify the Pattern
The numbers in the series are \(54, 60, 66, 72, 78, \ldots, 120\). We can see that these numbers are in an arithmetic progression (AP) with a common difference of \(6\).

### Step 2: Find the First and Last Terms
The first term \(a = 54\) and the last term \(l = 120\).

### Step 3: Determine the Number of Terms
To find the number of terms \(n\) in the series, we can use the formula for the \(n\)-th term of an AP:
\[ l = a + (n-1)d \]
where \(d\) is the common difference. Plugging in the values, we have:
\[ 120 = 54 + (n-1) \cdot 6 \]
\[ 120 - 54 = (n-1) \cdot 6 \]
\[ 66 = (n-1) \cdot 6 \]
\[ n - 1 = \frac{66}{6} = 11 \]
\[ n = 12 \]

### Step 4: Factor Out the Common Term
Next, we can factor out \(6^3\) from each term:
\[ 54^3 + 60^3 + 66^3 + 72^3 + 78^3 + \ldots + 120^3 = 6^3 \left(9^3 + 10^3 + 11^3 + \ldots + 20^3\right) \]

### Step 5: Sum of Cubes Formula
We need to calculate the sum \(9^3 + 10^3 + 11^3 + \ldots + 20^3\). The formula for the sum of cubes of the first \(n\) natural numbers is:
\[ \left(\frac{n(n+1)}{2}\right)^2 \]
To find the sum from \(9\) to \(20\), we can compute:
\[ \text{Sum from } 1 \text{ to } 20 - \text{Sum from } 1 \text{ to } 8 \]

### Step 6: Calculate the Sums
1. **Sum from 1 to 20**:
\[ \left(\frac{20 \cdot 21}{2}\right)^2 = (210)^2 = 44100 \]
2. **Sum from 1 to 8**:
\[ \left(\frac{8 \cdot 9}{2}\right)^2 = (36)^2 = 1296 \]
3. **Sum from 9 to 20**:
\[ 44100 - 1296 = 42804 \]

### Step 7: Final Calculation
Now we can substitute back into our equation:
\[ 54^3 + 60^3 + 66^3 + 72^3 + 78^3 + \ldots + 120^3 = 6^3 \cdot 42804 \]
Calculating \(6^3\):
\[ 6^3 = 216 \]
Thus:
\[ 216 \cdot 42804 = 9246654 \]

### Conclusion
The value of \(54^3 + 60^3 + 66^3 + 72^3 + 78^3 + \ldots + 120^3\) is **9246654**.