Vassil from Lawnswood School, Leeds, Michael from Madras College
St Andrews and Koopa Koo from Boston College all solved this problem,
well done all of you.
Here is Vassil's solution:
Let f(n) denote the sum of the first n terms of the sequence
0, 1, 1, 2, 2, 3, 3,¼, p, p, p+1, p+1,¼.
First I tried with several numbers.
Let n=15. Then f(15)=2*(1+2+3+4+5+6+7)=7*8
where the sequence is:
0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7
Let n=14. Then f(14)=2*(1+2+3+4+5+6+7)-7=7*7
where the sequence is:
0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7
Let n=17. Then f(17)=2*(1+2+3+4+5+6+7+8)=8*9
where the sequence is:
0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8
Let n=16. Then f(16)=2*(1+2+3+4+5+6+7+8)-8=8*8
where the sequence is:
0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8
I noticed that the formula for f(n) depends on whether n is odd or even.
Case I - n is odd, i.e. n=2k+1
Then
f(n)
= 2(1+2+...+k)
= 2k(k+1)/2
=
æ ç
è
n - 12
ö ÷
ø
æ ç
è
n+12
ö ÷
ø
.
Case II - n is even, i.e. n=2k
f(n)
= 2(1+2+...+k) - k
= k2 + k - k
= k2
=
æ ç
è
n2
ö ÷
ø
2
.
Now we have to calculate f(a+b)-f(a-b)
There are two cases. In the first case, when one of a and b is even and the other
is odd, then (a+b) and (a-b) are both odd. Otherwise
(a+b) and (a-b) are both even.
Case I (a+b) and (a-b) both odd.