Question

Do there exist consecutive odd integers a1,.....ak, and consecutive
even integers b1,...... bk (where k is greater than or equal to 1) such that
a1 +.......+ ak = b1 + ........+ bk ?
(The k and 1 are subscripts)

Answer 1

intuitively i think not - but proving it is something else!
suppose you pick out 3 of each which are close in value
2 + 4 + 6 = 12
1 + 3 + 5 = 9
3 + 5 + 7 = 15
etc - they will never be equal - they always be different by the total of numbers in each sequence

maybe the numbers either side of 0 might give equal sums?

Answer 2

i think not was well. maybe a proof by induction is necessary, but first off note that the sum of even numbers is always even. so k would have to be even.

then notice that if k = 2 you have
\[(2j-1)+(2j+1)=4j= 2\times 2j\] so it is twice an even number.
then notice that if k = 4 you can arrange them as
\[2j-3+2j-1+2j+1+2j+3=8j\] a number divisible by 8, whereas
\[2m-2+2m+2m+2+2m+4=8m+4\] which is not divisible by 8. i think you can extend this to k terms