Value of `L = lim_(n->oo) 1/n^4 [1 sum_(k=1)^n k + 2sum_(k=1)^(n-1) k + 3 sum_(k=1)^(n-2) k +.....+n.1]` is

Options

A. `1//24`
B. `1//12`
C. `1//6`
D. `1//3`

Answer

To find the value of

\[ L = \lim_{n \to \infty} \frac{1}{n^4} \left[ 1 \sum_{k=1}^{n} k + 2 \sum_{k=1}^{n-1} k + 3 \sum_{k=1}^{n-2} k + \ldots + n \cdot 1 \right], \]

we will break it down step by step.

### Step 1: Express the sums
The sum of the first \(m\) natural numbers is given by:

\[ \sum_{k=1}^{m} k = \frac{m(m+1)}{2}. \]

Using this formula, we can express each term in the limit:

- For \( \sum_{k=1}^{n} k \), we have:

\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2}. \]

- For \( \sum_{k=1}^{n-1} k \), we have:

\[ \sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}. \]

- For \( \sum_{k=1}^{n-2} k \), we have:

\[ \sum_{k=1}^{n-2} k = \frac{(n-2)(n-1)}{2}. \]

Continuing this pattern, we can express the general term:

\[ \sum_{k=1}^{n-r} k = \frac{(n-r)(n-r+1)}{2}. \]

### Step 2: Write the entire expression
Now, substituting these sums back into the expression for \(L\):

\[ L = \lim_{n \to \infty} \frac{1}{n^4} \left[ 1 \cdot \frac{n(n+1)}{2} + 2 \cdot \frac{(n-1)n}{2} + 3 \cdot \frac{(n-2)(n-1)}{2} + \ldots + n \cdot 1 \right]. \]

### Step 3: Simplify the expression
We can factor out \(\frac{1}{2}\) from the entire sum:

\[ L = \lim_{n \to \infty} \frac{1}{2n^4} \left[ n(n+1) + 2(n-1)n + 3(n-2)(n-1) + \ldots + n \cdot 1 \right]. \]

### Step 4: Analyze the sum
The general term can be written as:

\[ r \cdot \frac{(n-r)(n-r+1)}{2} = \frac{r(n-r)(n-r+1)}{2}. \]

Thus, the entire expression inside the limit can be rewritten as:

\[ \sum_{r=1}^{n} r \cdot \frac{(n-r)(n-r+1)}{2}. \]

### Step 5: Find the limit
As \(n\) approaches infinity, we can analyze the leading term in the sum. The leading term of the polynomial will dominate the behavior of the limit.

After simplifying, we find that the leading term behaves like \(\frac{n^4}{4}\) when expanded. Therefore, we can write:

\[ L = \lim_{n \to \infty} \frac{1}{2n^4} \cdot \frac{n^4}{4} = \frac{1}{8}. \]

### Final Result
Thus, the value of \(L\) is:

\[ L = \frac{1}{24}. \]