Alternating Squares
David W. Hansen
© 2004
A series in which the sign of each term alternates between positive and negative is called an
alternating series. 12 – 22 + 32 – 42 is an example of an alternating series of squares. Let’s look at
some alternating series of squares of consecutive natural numbers. Here are a few of them.
12 – 22 = 1 – 4 = - 3 = - (1 + 2)
12 – 22 + 32 = 1 – 4 + 9 = 6 = + (1 + 2 + 3)
12 – 22 + 32 – 42 = 1 – 4 + 9 – 16 = - 10 = - (1 + 2 + 3 + 4),
and in general, we have
12 – 22 + 32 – 42 + . . . + (-1)n+1n2 = (-1)n+1 (1 + 2 + 3 + . . . + n). (1)
We shall prove this by mathematical induction
1. Let n = 1. Then, 12 = 1, and (-1)1+1 = 1. Thus, the theorem (1) is true for n = 1. It is also true for
n = 2, 3, and 4 as shown above.
2. Assume (1) is true for n = k, where k is some natural number. Then, we know that
12 – 22 + 32 – 42 + . . . + (-1)k+1k2 = (-1)k+1 (1 + 2 + 3 + . . . + k) (2)
3. Let n = k+1. We must then prove that
12 – 22 + 32 – 42 + . . . + (-1)k+2(k+1)2 = (-1)k+2 (1 + 2 + 3 + . . . + k+1 ) (3)
4. Proof of (3). 12 – 22 + . . . + (-1)k+2(k+1)2
= [12 – 22 + . . . + (-1)k+1k2 ] + (-1)k+2(k+1)2
= (-1)k+1 (1 + 2 + . . . + k) + (-1)k+2(k+1)2 ( substituting from (2) above )
= (-1)k+1 [ k(k+1)/2 ] + (-1)k+2(k+1)2 ( since 1 + 2 + . . . + k = k(k+1)/2 )
= (-1)k+1 [ k(k+1)/2 ] + (-1)k+1(-1) (k+1)2 ( factoring out -1 from (-1)k+2 )
= (-1)k+1 (k+1) [ k/2 + (-1)2(k+1)/2 ( factoring out (-1)k+1 (k+1) and getting a
common denominator of 2 )
= (-1)k+1 (k+1) [ k - 2k - 2 ] / 2
= (-1)k+1 (k+1) [ - k - 2 ] / 2 = (-1)k+1 (k+1) (- 1) [k + 2] / 2 = (-1)k+2 (k+1)(k+2) / 2
= (-1)k+2 (1 + 2 + 3 + . . . + k+1 ), [ since 1 + 2 + 3 + . . . + k+1 = (k+1)(k+2)/2 ],
which is (3) above.
5. By the Principle of Mathematical Induction, the theorem (1) is proved.
Example. Find the value of 12 – 22 + 32 – 42 + 52 – 62 + 72 – 82 without squaring any numbers. Since
n = 8, the value is (-1)9 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = – 36.