`lim_(ktooo)(1^(3)+2^(3)+……….+k^(3))/(k^(4))` is equal to

Options

A. A. 0
B. B. 2
C. C. `1/3`
D. D. `1/4`

Answer

To solve the limit

\[ \lim_{k \to \infty} \frac{1^3 + 2^3 + \ldots + k^3}{k^4}, \]

we can follow these steps:

### Step 1: Use the formula for the sum of cubes
The sum of the first \( k \) cubes is given by the formula:

\[ 1^3 + 2^3 + \ldots + k^3 = \left( \frac{k(k + 1)}{2} \right)^2. \]

### Step 2: Substitute the sum into the limit
Substituting this formula into our limit, we have:

\[ \lim_{k \to \infty} \frac{\left( \frac{k(k + 1)}{2} \right)^2}{k^4}. \]

### Step 3: Simplify the expression
Now we simplify the expression:

\[ = \lim_{k \to \infty} \frac{k^2(k + 1)^2}{4k^4}. \]

### Step 4: Factor out \( k^4 \)
We can factor out \( k^4 \) from the denominator:

\[ = \lim_{k \to \infty} \frac{(k^2)(k^2 + 2k + 1)}{4k^4} = \lim_{k \to \infty} \frac{(k^2 + 2k + 1)}{4k^2}. \]

### Step 5: Divide by \( k^2 \)
Now we divide each term in the numerator by \( k^2 \):

\[ = \lim_{k \to \infty} \frac{1 + \frac{2}{k} + \frac{1}{k^2}}{4}. \]

### Step 6: Evaluate the limit
As \( k \) approaches infinity, \( \frac{2}{k} \) and \( \frac{1}{k^2} \) approach 0:

\[ = \frac{1 + 0 + 0}{4} = \frac{1}{4}. \]

Thus, the final answer is:

\[ \lim_{k \to \infty} \frac{1^3 + 2^3 + \ldots + k^3}{k^4} = \frac{1}{4}. \]

### Conclusion
The limit evaluates to \( \frac{1}{4} \).