Question 2: Sum of squares over sum of numbers

Algebra Level 3

What most specific condition necessarily applies to $k$ such that

$\large \frac{1+4+9+16+...+k^2}{1+2+3+4+...+k} = C$

where $C$ is an integer.

k \equiv 0 \text{ (mod 3)}

k \equiv 2 \text{ (mod 3)}

k \equiv 1 \text{ (mod 6)}

k \equiv 1 \text{ (mod 3)}

1 solution

Edwin Gray
Aug 16, 2018

The sum of k squares is given by k(k + 1)(2k + 1)/6. The sum of numbers from 1 to k is given by k(k + 1)/ 2. Then C is the ratio = k(k + 1)(2k + 1)*2/6k(k + 1) which is equal to (2k + 1)/2. So the condition is 2k + 1. cong. 0 (mod 3),which has the solution k = 1,4,7,..... So k.cong. 1 (mod 3). Ed Gray

Question 2: Sum of squares over sum of numbers

1 solution

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