Question

If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?

Answer 1

If
`k` is odd, then


`\displaystyle\sum_(n=1)^(\lfloor k/2\rfloor)2n=2\frac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\frac k2\right\rfloor^2+\left\lfloor\frac k2\right\rfloor`

while if
`k` is even, then the sum would be


`\displaystyle\sum_(n=1)^(k/2)2n=2\frac{\frac k2\left(\frac k2+1\right)}2=\frac{k^2+2k}4`

The latter case is easier to solve:


`\frac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0`

which means
`k=6`.

In the odd case, instead of considering the above equation we can consider the partial sums. If
`k` is odd, then the sum of the even integers between 1 and
`k` would be


`S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)`

Now consider the partial sum up to the second-to-last term,


`S^*=2+4+6+\cdots+(k-5)+(k-3)`

Subtracting this from the previous partial sum, we have


`S-S^*=k-1`

We're given that the sums must add to
`2k`, which means


`S=2k`

`S^*=2(k-2)`

But taking the differences now yields


`S-S^*=2k-2(k-2)=4`

and there is only one
`k` for which
`k-1=4`; namely,
`k=5`. However, the sum of the even integers between 1 and 5 is
`2+4=6`, whereas
`2k=10\\eq6`. So there are no solutions to this over the odd integers.