Consider the sequence in the form of group (1),(2,2)(3,3,3),(4,4,4,4),(5,5,5,5,5…..)

Options

A. `a/x+c/y=2`
B. `a/x+c/y=c/a`
C. `1/x +1/y=2/b`
D. `1/x+1/y=2/ac`

Answer

To find the sum of the first 2000 terms of the given sequence, we can break down the problem step by step.

### Step 1: Understand the Sequence
The sequence is formed by groups:
- 1st group: (1)
- 2nd group: (2, 2)
- 3rd group: (3, 3, 3)
- 4th group: (4, 4, 4, 4)
- ...
- nth group: (n, n, n, ..., n) (n times)

The total number of terms in the first n groups can be calculated using the formula for the sum of the first n natural numbers, which is given by:
\[ S_n = \frac{n(n + 1)}{2} \]

### Step 2: Determine Which Group Contains the 2000th Term
We need to find the largest integer \( n \) such that:
\[ S_n \leq 2000 \]
This can be expressed as:
\[ \frac{n(n + 1)}{2} \leq 2000 \]
Multiplying both sides by 2:
\[ n(n + 1) \leq 4000 \]

### Step 3: Solve the Inequality
We need to find the largest \( n \) satisfying \( n(n + 1) \leq 4000 \).
- Testing \( n = 63 \):
\[ 63 \times 64 = 4032 \quad (\text{too large}) \]
- Testing \( n = 62 \):
\[ 62 \times 63 = 3906 \quad (\text{valid}) \]

Thus, \( n \) can be at most 62. Therefore, the 2000th term falls in the 63rd group.

### Step 4: Calculate the Total Number of Terms Up to the 62nd Group
The total number of terms in the first 62 groups is:
\[ S_{62} = \frac{62 \times 63}{2} = 1953 \]

### Step 5: Determine the Number of Terms in the 63rd Group
The number of terms from the 1954th to the 2000th term is:
\[ 2000 - 1953 = 47 \]
Thus, the 63rd group contributes 47 terms of 63.

### Step 6: Calculate the Sum of the First 2000 Terms
The sum of the first 2000 terms can be calculated as:
\[ \text{Sum} = \text{Sum of terms from groups 1 to 62} + \text{Sum of terms in the 63rd group} \]
The sum of squares of the first 62 groups is:
\[ \sum_{k=1}^{62} k^2 = \frac{62(63)(125)}{6} \]
Calculating this gives:
\[ \sum_{k=1}^{62} k^2 = 16285 \]

The sum of the 47 terms in the 63rd group is:
\[ 63 \times 47 = 2961 \]

### Final Calculation
Thus, the total sum of the first 2000 terms is:
\[ \text{Total Sum} = 16285 + 2961 = 19246 \]

### Conclusion
The sum of the first 2000 terms of the sequence is **19246**.

---