Let S_{n} = sum k = 1 to 4n (- 1) ^ ((k(k 1))/2) * k ^ 2 Then S_{n} can take value(s

Key Ideas

1. Alternating sign patterns: The factor $(-1)^{\frac{k(k+1)}{2}}$ produces a 4-term cycle of signs. Matching the cycle length with the summation limits is a classic trick.
2. Block summation: Whenever the sign (or any function) repeats after a small number $p$, it is efficient to group terms in blocks of size $p$. Each block often simplifies nicely.
3. Arithmetic series of coefficients: After simplifying one block we often get an expression like $a m + b$ (linear in the block index $m$). Summing linear expressions over $m$ is straightforward using the formula for $\sum m$.

Logical Chain of Thought

1. Write the sum

   $$S_n = \sum_{k=1}^{4n} (-1)^{\frac{k(k+1)}{2}}\,k^{2}$$

2. Find the sign for $k = 1,2,3,4$ to see the 4-term cycle: $[-1,-1,+1,+1]$.

3. Set $k = 4m+1,\,4m+2,\,4m+3,\,4m+4$ for the $(m+1)$-th block ($m=0,1,\dots,n-1$).

4. Compute the block sum (lots of cancellation) and obtain $32m + 20$.

5. Sum $32m + 20$ for $m=0$ to $n-1$ using

   $$\sum_{m=0}^{n-1} m = \frac{n(n-1)}{2}.$$

6. Simplify to get

   $$S_n = 16n^{2} + 4n = 4n(4n+1).$$

7. Conclude that for any positive integer $n$, $S_n$ always lands on the values $4n(4n+1)$ (and nothing else).

What is the question?

We have a long addition (a series) where the sign in front of each square number keeps switching in a fixed pattern: minus, minus, plus, plus. We add together the first $4n$ square numbers following that sign pattern.

How do we tackle it?

1. Notice the pattern: Every 4 numbers repeat the same signs.
2. Group in blocks of 4: Because the pattern repeats exactly after 4 terms, add the terms 4 at a time.
3. Find the sum of one block: Work out the total for any block. (Good news—most parts cancel!)
4. Add all blocks together: There are $n$ such blocks, so multiply the block–sum by $n$, then do a tiny extra tidy-up.

In the end you discover the whole big sum is really neat: it always equals $4n(4n+1)$.

Step 1: Identify the sign pattern

For $k=1,2,3,4$:

Hence the signs repeat every 4 terms: $[-1,-1,+1,+1]$.

Step 2: Group terms in blocks of 4

Write $k = 4m + r$ where $r = 1,2,3,4$ and $m = 0,1,\dots,n-1$.

Step 3: Compute the sum for one block

Step 4: Sum all $n$ blocks

Using $\sum_{m=0}^{n-1} m = \frac{n(n-1)}{2}$,

Final Answer