WWW.C E M C .U WAT E R LO O.C A | T h e C E N T R E fo r E D U C AT I O N i n M AT H E M AT I C S a n d CO M P U T I N G
Problem of the Week
Problem E and Solution
A Curious Pattern
Problem
Let S(n) be the sum of the first n terms of the series
1 − 2 + 3 − 4 + 5 − 6 + 7 − 8 + 9 − 10 + 11 − 12 + . . .
Show that S(a) + S(b) + S(a + b) = 1 when both a and b are odd.
Solution
a and b are both odd.
Let a = 2k − 1 and b = 2m − 1, for some positive integers k and m. If k = m then a = b but we
want to prove the result true for any odd positive integers, a and b.
We have S(a) = S(2k − 1) = k and S(b) = S(2m − 1) = m.
Also, a + b = (2k − 1) + (2m − 1) = 2k + 2m − 2 = 2(k + m − 1) is even, so
S(a + b) = S(2(k + m − 1)) = −(k + m − 1) = −k − m + 1.
Then S(a) + S(b) + S(a + b) = k + m + (−k − m + 1) = 1.
Therefore, S(a) + S(b) + S(a + b) = 1 whenever a and b are odd positive integers.
Extension
Since we have formulas for S(a) and S(b) based on whether a and b are even or odd, we will
break into cases based on the parity of a and b to show that none of the other possible cases
make S(a) + S(b) + S(a + b) = 1 true.
Case 1: a and b are both even
Let a = 2k and b = 2m, for some positive integers k and m.
We have S(a) = S(2k) = −k and S(b) = S(2m) = −m.
Also, a + b = 2k + 2m = 2(k + m) is even, so S(a + b) = S(2(k + m)) = −(k + m) = −k − m.
The equation S(a) + S(b) + S(a + b) = 1 becomes −k − m + (−k − m) = 1 or 2k + 2m = −1.
Since k and m are positive integers, 2k + 2m is also a positive integer so we cannot have
2k + 2m = −1. Therefore, there is no solution.
Case 2: a is even and b is odd
Let a = 2k and b = 2m − 1, for some positive integers k and m.
We have S(a) = S(2k) = −k and S(b) = S(2m − 1) = m.
Also, a + b = 2k + (2m − 1) = 2(k + m) − 1 is odd, so S(a + b) = S(2(k + m) − 1) = k + m.
The equation S(a) + S(b) + S(a + b) = 1 becomes −k + m + (k + m) = 1 or 2m = 1, thus
m = 12 . But m must an integer, so there is no solution.
Case 3: a is odd and b is even
Let a = 2k − 1 and b = 2m, for some positive integers k and m.
We have S(a) = S(2k − 1) = k and S(b) = S(2m) = −m.
Also, a + b = (2k − 1) + 2m = 2(k + m) − 1 is odd, so S(a + b) = S(2(k + m) − 1) = k + m.
The equation S(a) + S(b) + S(a + b) = 1 becomes k − m + (k + m) = 1 or 2k = 1, thus k = 12 .
But k must be an integer, so there is no solution.
Therefore, S(a) + S(b) + S(a + b) = 1 is only true when both a and b are odd positive integers.
WWW.C E M C .U WAT E R LO O.C A | T h e C E N T R E fo r E D U C AT I O N i n M AT H E M AT I C S a n d CO M P U T I N G
We will now prove the first fact given in the question.
We will show that for numbers of the form n = 2k, we have S(n) = S(2k) = −k.
Numbers of the form n = 2k:
S(2k) = 1 − 2 + 3 − 4 + 5 + . . . + (2k − 1) − 2k
= (1 + 3 + 5 + . . . + (2k − 1)) − (2 + 4 + 6 + . . . + 2k)
grouping even and odd numbers
= (1 + 3 + 5 + . . . + (2k − 1)) − 2(1 + 2 + 3 + . . . + k)
factor out a 2 from the even numbers
= (1 + 3 + 5 + . . . + (2k − 1)) + (2 + 4 + 6 + . . . + 2k) − (2 + 4 + 6 + . . . + 2k) − 2(1 + 2 + 3 + . . . + k)
add and subtract the even numbers from 2 to 2k
= (1 + 2 + 3 + 4 + 5 + . . . + (2k − 1) + 2k) − (2 + 4 + 6 + . . . + 2k) − 2(1 + 2 + 3 + . . . + k)
= (1 + 2 + 3 + 4 + 5 + . . . + (2k − 1) + 2k) − 2(1 + 2 + 3 + . . . + k) − 2(1 + 2 + 3 + . . . + k)
= (1 + 2 + 3 + 4 + 5 + . . . + (2k − 1) + 2k) − 4(1 + 2 + 3 + . . . + k)
Now, using the formula for the sum of the first n integers, that is
1 + 2 + 3 + . . . + (n − 1) + n =
n(n + 1)
2
we have
S(2k) = (1 + 2 + 3 + 4 + 5 + . . . + (2k − 1) + 2k) − 4(1 + 2 + 3 + . . . + k)
(2k)(2k + 1)
k(k + 1)
=
−4
2
2
= k(2k + 1) − 2k(k + 1)
= 2k 2 + k − 2k 2 − 2k
= −k
The proof that for numbers of the form n = 2k − 1, we have S(n) = S(2k − 1) = k is similar.
We’ll leave that for you to try on your own.