1998 AIME Problem 5

Question

Given that

A_k = \frac{k(k - 1)}{2}\cos\frac{k(k - 1)\pi}{2},

find

|A_{19} + A_{20} + \cdots + A_{98}|.

Answer

\frac{k(k-1)}{2}

is an integer, so each cosine is

\pm 1,

with sign depending only on

k \bmod 4

Group the

80

terms into consecutive blocks of four starting at

k = 19;

consecutive triangular numbers differ by

k,

so each block collapses

Since

k(k-1)

is even,

n_k = \frac{k(k-1)}{2}

is an integer and

\cos\frac{k(k-1)\pi}{2}

= \cos(n_k \pi)

= (-1)^{n_k}.

The parity of the triangular number

n_k

depends only on

k \bmod 4:

it is even for

k \equiv 0, 1 \pmod 4

and odd for

k \equiv 2, 3 \pmod 4.

So

A_k = n_k

when

k \equiv 0, 1 \pmod 4

and

A_k = -n_k

when

k \equiv 2, 3 \pmod 4.

Group the

80

terms into

20

consecutive blocks of four starting at

k = 19 \equiv 3 \pmod 4.

Using

n_{j+1} - n_j = j,

each block with

k \equiv 3 \pmod 4

collapses:

\begin{aligned} &A_k + A_{k+1} \\ &\quad {}+ A_{k+2} + A_{k+3} \\ &= (n_{k+1} - n_k) \\ &\quad {}- (n_{k+3} - n_{k+2}) \\ &= k - (k + 2) \\ &= -2. \end{aligned}

The total is

20 \cdot (-2) = -40,

so the requested absolute value is

40.

Problem 5 in other years