Solved: "How to evaluate the following sum?∑i=1n[3n−i2]"

Adrienne Rowe

Answered question

2022-02-24

How to evaluate the following sum?

\sum_{i = 1}^{n} [\frac{3 n - i}{2}]

Goodwin2ug

Beginner2022-02-25Added 6 answers

This answer is only for n even. For n odd, it can be done similarly.
Write

n = 2 k

for some integer k. Then

\sum_{i = 1}^{n} [\frac{3 n - i}{2}] = \sum_{i = 1}^{2 k} [\frac{6 k - i}{2}]

= \sum_{i e v e n}^{2 k} [\frac{6 k - i}{2}] + \sum_{i o d d}^{2 k} [\frac{6 k - i}{2}]

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

= \sum_{i = 1}^{k} [3 k - i] + \sum_{i = 1}^{k} [3 k - i + \frac{1}{2}]

= \sum_{i = 1}^{k} (3 k - i) + \sum_{i = 1}^{k} (3 k - i)

= 2 \sum_{i = 1}^{k} (3 k - i)

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