Sequences & Series Question - None

Question

For Problems 13–15: Consider the sequence in the form of groups $(1), (2, 2), (3, 3, 3), (4, 4, 4, 4), (5, 5, 5, 5, 5), \ldots$

The sum of the remaining terms in the group after 2000^th term in which 2000^th term lies is

Accepted Answer

Step 1: Find which group contains the 2000th term.

Group n contains n terms. Total terms up to and including group n = 1 + 2 + 3 + ... + n = n(n+1)/2

We need n(n+1)/2 ≥ 2000

n(n+1) ≥ 4000

Testing: 63 × 64 = 4032 ✓ and 62 × 63 = 3906 ✗

So the 2000th term lies in group 63.

Step 2: Find the position of the 2000th term within group 63.

Terms up to end of group 62 = 62 × 63 / 2 = 1953

Position of 2000th term in group 63 = 2000 - 1953 = 47

Step 3: Calculate the sum of remaining terms in group 63.

Group 63 has 63 terms, each equal to 63

Sum of all terms in group 63 = 63 × 63 = 3969

Sum of first 47 terms in group 63 = 47 × 63 = 2961

Sum of remaining terms = 3969 - 2961 = 1008

Alternatively: Remaining terms = (63 - 47) × 63 = 16 × 63 = 1008

∴ Answer: C

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