The arithmetic mean of 1. 8. 27 64.. upto n terms is given by y

Options

A. `(n(n+1))/(2)`
B. `n(n+1)^(2)/(2)`
C. `n(n+1)^(2)/(4)`
D. `(n^(2)(n+1))^(2)/(4)`

Answer

To find the arithmetic mean of the sequence 1, 8, 27, 64 up to n terms, we can follow these steps:

### Step 1: Identify the Pattern
The numbers given are:
- 1 = \(1^3\)
- 8 = \(2^3\)
- 27 = \(3^3\)
- 64 = \(4^3\)

From this, we can see that the sequence consists of the cubes of the first n natural numbers.

### Step 2: Write the Formula for the Sum of Cubes
The sum of the cubes of the first n natural numbers can be expressed using the formula:
\[ \text{Sum} = \left(\frac{n(n + 1)}{2}\right)^2 \]
This formula states that the sum of the cubes of the first n natural numbers is equal to the square of the sum of the first n natural numbers.

### Step 3: Calculate the Arithmetic Mean
The arithmetic mean (y) is calculated by dividing the sum of the terms by the number of terms (n):
\[ y = \frac{\text{Sum of cubes}}{n} \]
Substituting the formula for the sum of cubes:
\[ y = \frac{\left(\frac{n(n + 1)}{2}\right)^2}{n} \]

### Step 4: Simplify the Expression
Now, we simplify the expression:
\[ y = \frac{n(n + 1)^2}{4} \]

### Final Answer
Thus, the arithmetic mean of the sequence 1, 8, 27, 64 up to n terms is:
\[ y = \frac{n(n + 1)^2}{4} \]

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