Consider the sequence in the form of group (1),(2,2)(3,3,3),(4,4,4,4),(5,5,5,5,5…..)
The sum of first 2000 terms is

Options

A. 7
B. 8
C. 9
D. 10

Answer

To solve the problem of finding the sum of the first 2000 terms of the given sequence, we can follow these steps:

### Step 1: Understand the Grouping of the Sequence
The sequence is grouped as follows:
- Group 1: (1)
- Group 2: (2, 2)
- Group 3: (3, 3, 3)
- Group 4: (4, 4, 4, 4)
- Group 5: (5, 5, 5, 5, 5)
- ...
- Group n: (n, n, n, ..., n) (n times)

### Step 2: Determine the Total Number of Terms in Each Group
The total number of terms up to group n is given by the formula:
\[ \text{Total terms} = \frac{n(n + 1)}{2} \]
This is the sum of the first n natural numbers.

### Step 3: Find the Group that Contains the 2000th Term
We need to find the largest integer n such that:
\[ \frac{n(n + 1)}{2} \leq 2000 \]
This can be rearranged to:
\[ n(n + 1) \leq 4000 \]

### Step 4: Solve the Inequality
We can solve for n by testing values:
- For \( n = 62 \):
\[ 62 \times 63 = 3906 \quad (\text{valid}) \]
- For \( n = 63 \):
\[ 63 \times 64 = 4032 \quad (\text{invalid}) \]

Thus, the largest n that satisfies the condition is \( n = 62 \).

### Step 5: Calculate the Total Number of Terms Up to the 62nd Group
The total number of terms up to the 62nd group is:
\[ \frac{62 \times 63}{2} = 1953 \]

### Step 6: Determine the 2000th Term
Since the 2000th term is in the 63rd group, we need to find how many terms are needed to reach 2000:
\[ 2000 - 1953 = 47 \]
Thus, the 2000th term is the 47th term in the 63rd group, which is 63.

### Step 7: Calculate the Sum of the First 2000 Terms
The sum of the first 2000 terms can be calculated as follows:
1. Sum of squares of the first 62 groups:
\[ \sum_{k=1}^{62} k^2 = \frac{62 \times 63 \times 125}{6} \]
(Using the formula for the sum of squares)

2. Add the contribution from the 63rd group:
\[ 63 \times 47 \]

### Step 8: Final Calculation
Calculating the sums:
1. Sum of squares:
\[ \frac{62 \times 63 \times 125}{6} = 162250 \]

2. Contribution from the 63rd group:
\[ 63 \times 47 = 2961 \]

3. Total sum:
\[ 162250 + 2961 = 165211 \]

Thus, the sum of the first 2000 terms is **165211**.