Daryl first writes the perfect squares as a sequence

$$
1,4,9,16,25,36,49,64,81,100, \ldots
$$

After the number 1, he then alternates by making two terms negative followed by leaving two terms positive. Daryl's new sequence is

$$
1,-4,-9,16,25,-36,-49,64,81,-100, \ldots
$$

What is the sum of the first 2011 terms in this new sequence?
(A) -4042109
(D) -4044121
(B) -4047638
(E) -4046132
(C) -4038094
The given sequence allows for many different patterns to be discovered depending on how terms in the sequence are grouped and then added.

One possibility is to add groups of four consecutive terms in the sequence.

That is, consider finding the sum of the sequence, $S$, in the manner shown below.

$$
S=(1+(-4)+(-9)+16)+(25+(-36)+(-49)+64)+(81+(-100)+(-121)+144)+\ldots
$$

The pattern that appears when grouping terms in this way is that each consecutive group of 4 terms, beginning at the first term, adds to 4 .

That is, $1+(-4)+(-9)+16=4,25+(-36)+(-49)+64=4,81+(-100)+(-121)+144=4$, and so on.

For now, we will assume that this pattern of four consecutive terms adding to 4 continues and wait to verify this at the end of the solution.

Since each consecutive group of four terms adds to 4 , the first eight terms add to 8 , the first twelve terms add to 12 and the first $n$ terms add to $n$ provided that $n$ is a multiple of 4 .

Thus, the sum of the first 2012 terms is 2012, since 2012 is a multiple of 4.

Since we are required to find the sum of the first 2011 terms, we must subtract the value of the $2012^{\text {th }}$ term from our total of 2012 .

We know that the $n^{\text {th }}$ term in the sequence is either $n^{2}$ or it is $-n^{2}$.

Therefore, we must determine if the $2012^{\text {th }}$ term is positive or negative.

By the alternating pattern of the signs, the first and fourth terms in each of the consecutive groupings will be positive, while the second and third terms are negative.

Since the $2012^{\text {th }}$ term is fourth in its group of four, its sign is positive.

Thus, the $2012^{\text {th }}$ term is $2012^{2}$.

Therefore, the sum of the first 2011 terms is the sum of the first 2012 terms, which is 2012, less the $2012^{\text {th }}$ term, which is $2012^{2}$.

Thus, $S=2012-2012^{2}=2012-4048144=-4046132$.

## Verifying the Pattern

While we do not expect that students will verify this pattern in the context of a multiple choice contest, it is always important to verify patterns.

One way to verify that the sum of each group of four consecutive terms (beginning with the first term) adds to 4 , is to use algebra.

If the first of four consecutive integers is $n$, then the next three integers in order are $n+1$, $n+2$, and $n+3$.

Since the terms in our sequence are the squares of consecutive integers, we let $n^{2}$ represent the first term in the group of four.

The square of the next integer larger than $n$ is $(n+1)^{2}$, and thus the remaining two terms in the group are $(n+2)^{2}$ and $(n+3)^{2}$.

Since the first and fourth terms are positive, while the second and third terms are negative, the sum of the four terms is $n^{2}-(n+1)^{2}-(n+2)^{2}+(n+3)^{2}$.

(For example, if $n=5$ then we have $5^{2}-6^{2}-7^{2}+8^{2}$.)

To simplify this expression, we must first understand how to simplify its individual parts such as $(n+1)^{2}$.

The expression $(n+1)^{2}$ means $(n+1) \times(n+1)$.

To simplify this product, we multiply the $n$ in the first set of brackets by each term in the second set of brackets and do the same for the 1 appearing in the first set of brackets.

The operation between each product remains as it appears in the expression, as an addition.

That is,

$$
\begin{aligned}
(n+1)^{2} & =(n+1) \times(n+1) \\
& =n \times n+n \times 1+1 \times n+1 \times 1 \\
& =n^{2}+n+n+1 \\
& =n^{2}+2 n+1
\end{aligned}
$$

Applying this process again,

$(n+2)^{2}=(n+2) \times(n+2)=n \times n+n \times 2+2 \times n+2 \times 2=n^{2}+2 n+2 n+4=n^{2}+4 n+4$, and

$(n+3)^{2}=(n+3) \times(n+3)=n \times n+n \times 3+3 \times n+3 \times 3=n^{2}+3 n+3 n+9=n^{2}+6 n+9$. Therefore, the sum of each group of four consecutive terms (beginning with the first term) is,

$$
\begin{aligned}
n^{2}-(n+1)^{2}-(n+2)^{2}+(n+3)^{2} & =n^{2}-\left(n^{2}+2 n+1\right)-\left(n^{2}+4 n+4\right)+\left(n^{2}+6 n+9\right) \\
& =n^{2}-n^{2}-2 n-1-n^{2}-4 n-4+n^{2}+6 n+9 \\
& =n^{2}-n^{2}-n^{2}+n^{2}-2 n-4 n+6 n+9-1-4 \\
& =4
\end{aligned}
$$

ANswer: (E)