The limit of `(1)/(n ^(4)) sum _(k =1) ^(n) k (k +2) (k +4) as n to oo` is equal to `(1)/(lamda),` then `lamda =`

Answer

To solve the limit of the expression \(\frac{1}{n^4} \sum_{k=1}^{n} k(k+2)(k+4)\) as \(n\) approaches infinity, we will follow these steps:

### Step 1: Expand the expression inside the summation
We start by expanding \(k(k+2)(k+4)\):
\[ k(k+2)(k+4) = k(k^2 + 6k + 8) = k^3 + 6k^2 + 8k \]

### Step 2: Write the summation
Now we can rewrite the summation:
\[ \sum_{k=1}^{n} k(k+2)(k+4) = \sum_{k=1}^{n} (k^3 + 6k^2 + 8k) \]
This can be separated into three different summations:
\[ \sum_{k=1}^{n} k^3 + 6\sum_{k=1}^{n} k^2 + 8\sum_{k=1}^{n} k \]

### Step 3: Use formulas for summations
We will use the formulas for the sums of powers of integers:
- \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\)
- \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\)
- \(\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2\)

Substituting these formulas into our expression:
\[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 \]
\[ 6\sum_{k=1}^{n} k^2 = 6 \cdot \frac{n(n+1)(2n+1)}{6} = n(n+1)(2n+1) \]
\[ 8\sum_{k=1}^{n} k = 8 \cdot \frac{n(n+1)}{2} = 4n(n+1) \]

### Step 4: Combine the results
Now, we combine all these results:
\[ \sum_{k=1}^{n} k(k+2)(k+4) = \left(\frac{n(n+1)}{2}\right)^2 + n(n+1)(2n+1) + 4n(n+1) \]

### Step 5: Simplify the expression
Now we need to simplify this expression. The leading term will dominate as \(n\) approaches infinity:
\[ \left(\frac{n(n+1)}{2}\right)^2 = \frac{n^2(n+1)^2}{4} \approx \frac{n^4}{4} \]
The other terms will grow slower than \(n^4\) and can be ignored in the limit:
\[ n(n+1)(2n+1) \approx 2n^3 \quad \text{and} \quad 4n(n+1) \approx 4n^2 \]

### Step 6: Substitute back into the limit
Now substituting back into the limit:
\[ \lim_{n \to \infty} \frac{1}{n^4} \left( \frac{n^4}{4} + \text{lower order terms} \right) = \lim_{n \to \infty} \left( \frac{1}{4} + \frac{\text{lower order terms}}{n^4} \right) \]
As \(n\) approaches infinity, the lower order terms divided by \(n^4\) will approach 0. Thus, we have:
\[ \lim_{n \to \infty} \frac{1}{n^4} \sum_{k=1}^{n} k(k+2)(k+4) = \frac{1}{4} \]

### Step 7: Relate to \(\lambda\)
We are given that this limit is equal to \(\frac{1}{\lambda}\), so:
\[ \frac{1}{\lambda} = \frac{1}{4} \implies \lambda = 4 \]

### Final Answer
Thus, the value of \(\lambda\) is:
\[ \lambda = 4 \]