Let P = 1 + 3 + 5 + … + (2k − 1) be the sum of the odd integers from 1 to 2k,…
2008
Let P = 1 + 3 + 5 + … + (2k − 1) be the sum of the odd integers from 1 to 2k, and let Q = 2 + 4 + 6 + … + 2k be the sum of the even integers from 1 to 2k, where k is a positive integer. Then:
- A.
P = Q - k
- B.
P = Q + k
- C.
P = Q
- D.
P = Q +2 k
Attempted by 18 students.
Show answer & explanation
Correct answer: A
Solution: Find each sum and compare.
Compute P (sum of odd integers from 1 to 2k): P = 1 + 3 + ... + (2k - 1) = k^2.
Compute Q (sum of even integers from 1 to 2k): Q = 2 + 4 + ... + 2k = 2(1 + 2 + ... + k) = 2 * k(k + 1)/2 = k(k + 1).
Compare: Q - P = k(k + 1) - k^2 = k. Therefore P = Q - k.
Quick check: for k = 3, P = 1 + 3 + 5 = 9 and Q = 2 + 4 + 6 = 12, so P = Q - 3 holds.