Let `lim_(n->oo)sum_(k=1)^n(lambdak^4+2k^3+k^2+k+1)/(3n^5+n^2+n+5k)=1/3` then `lambda` is equal to

Explanation

To solve the limit problem given, we will follow these steps:

### Step 1: Write the limit expression
We start with the limit expression:
\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{\lambda k^4 + 2k^3 + k^2 + k + 1}{3n^5 + n^2 + n + 5k} \]

### Step 2: Simplify the denominator
As \( n \) approaches infinity, the dominant term in the denominator is \( 3n^5 \). Thus, we can factor out \( 3n^5 \):
\[ 3n^5 + n^2 + n + 5k \approx 3n^5 \left(1 + \frac{n^2}{3n^5} + \frac{n}{3n^5} + \frac{5k}{3n^5}\right) \]
This simplifies to:
\[ 3n^5 \left(1 + \frac{1}{3n^3} + \frac{1}{3n^4} + \frac{5k}{3n^5}\right) \]

### Step 3: Rewrite the limit
Now we can rewrite the limit as:
\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{\lambda k^4 + 2k^3 + k^2 + k + 1}{3n^5} \cdot \frac{1}{1 + \frac{1}{3n^3} + \frac{1}{3n^4} + \frac{5k}{3n^5}} \]

### Step 4: Evaluate the sum
The sum can be approximated as \( n \) approaches infinity. The dominant term in the numerator is \( \lambda k^4 \), so we can focus on that:
\[ \sum_{k=1}^{n} k^4 \approx \frac{n^5}{5} \]
Thus, we have:
\[ \sum_{k=1}^{n} \left(\lambda k^4 + 2k^3 + k^2 + k + 1\right) \approx \lambda \cdot \frac{n^5}{5} + 2 \cdot \frac{n^4}{4} + \frac{n^3}{3} + \frac{n^2}{2} + n \]

### Step 5: Substitute back into the limit
Now substituting back into the limit:
\[ \lim_{n \to \infty} \frac{\lambda \cdot \frac{n^5}{5} + O(n^4)}{3n^5} \]
This simplifies to:
\[ \lim_{n \to \infty} \left(\frac{\lambda}{15} + O\left(\frac{1}{n}\right)\right) = \frac{\lambda}{15} \]

### Step 6: Set the limit equal to \( \frac{1}{3} \)
According to the problem, this limit equals \( \frac{1}{3} \):
\[ \frac{\lambda}{15} = \frac{1}{3} \]

### Step 7: Solve for \( \lambda \)
Multiplying both sides by 15 gives:
\[ \lambda = 5 \]

### Final Answer
Thus, the value of \( \lambda \) is:
\[ \lambda = 5 \]

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