`lim_(n->oo) (1.2+2.3+3.4+....+n(n+1))/n^3`

Options

A. 1
B. -1
C. `1/3`
D. None of these

Answer

To solve the limit

\[ \lim_{n \to \infty} \frac{1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + n(n+1)}{n^3}, \]

we first need to simplify the numerator, which is the sum of the series \(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + n(n+1)\).

### Step 1: Rewrite the Series

The term \(k(k+1)\) can be rewritten as \(k^2 + k\). Thus, we can express the series as:

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

### Step 2: Use Known Formulas for Sums

We can use the formulas for the sums of the first \(n\) natural numbers and the sum of the squares of the first \(n\) natural numbers:

- The sum of the first \(n\) natural numbers is given by:

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

- The sum of the squares of the first \(n\) natural numbers is given by:

\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}. \]

### Step 3: Substitute the Formulas

Substituting these formulas into our expression, we have:

\[ \sum_{k=1}^{n} k(k+1) = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}. \]

### Step 4: Combine the Terms

To combine these two fractions, we need a common denominator. The common denominator is 6:

\[ \sum_{k=1}^{n} k(k+1) = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6} = \frac{n(n+1)(2n+1 + 3)}{6} = \frac{n(n+1)(2n + 4)}{6}. \]

### Step 5: Simplify the Expression

Now we can simplify the expression:

\[ \sum_{k=1}^{n} k(k+1) = \frac{n(n+1)(2(n + 2))}{6} = \frac{n(n+1)(n + 2)}{3}. \]

### Step 6: Substitute Back into the Limit

Now substitute this back into the limit:

\[ \lim_{n \to \infty} \frac{\frac{n(n+1)(n+2)}{3}}{n^3} = \lim_{n \to \infty} \frac{n(n+1)(n+2)}{3n^3}. \]

### Step 7: Simplify the Limit

This simplifies to:

\[ \lim_{n \to \infty} \frac{(n+1)(n+2)}{3n^2}. \]

### Step 8: Divide Each Term by \(n\)

Dividing each term by \(n\):

\[ \lim_{n \to \infty} \frac{(1 + \frac{1}{n})(1 + \frac{2}{n})}{3} = \frac{1 \cdot 1}{3} = \frac{1}{3}. \]

### Final Answer

Thus, the limit is:

\[ \boxed{\frac{1}{3}}. \]