Question

Let Sn = 4n∑k=1(-1)k(k+1)/2K2. Then Sn can take value(s).

• A. 1056
• B. 1088
• C. 1120
• D. 1332

Answer 1

Answer: (A), (D)

1056, 1332

Answer 2

The given series is $S_n = \sum_{k=1}^{4n} (-1)^{k(k+1)/2} k^2$.

First, let's analyze the pattern of the exponent $k(k+1)/2 \pmod 2$. This is the $k$-th triangular number, $T_k$.

• For $k=1$, $T_1 = 1$. $(-1)^{T_1} = -1$.
• For $k=2$, $T_2 = 3$. $(-1)^{T_2} = -1$.
• For $k=3$, $T_3 = 6$. $(-1)^{T_3} = +1$.
• For $k=4$, $T_4 = 10$. $(-1)^{T_4} = +1$.

The pattern of signs for $(-1)^{k(k+1)/2}$ repeats every four terms: $(-1, -1, +1, +1)$.

Now, let's group the terms of the series in blocks of four. Let $j$ be the block index, starting from $j=0$. The terms in the $j$-th block are for $k = 4j+1, 4j+2, 4j+3, 4j+4$.

The sum of a block of four terms, $P_j$, is: $P_j = -(4j+1)^2 - (4j+2)^2 + (4j+3)^2 + (4j+4)^2$

We can rearrange this as: $P_j = [(4j+4)^2 - (4j+2)^2] + [(4j+3)^2 - (4j+1)^2]$

Using the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$:

$(4j+4)^2 - (4j+2)^2 = ((4j+4)-(4j+2))((4j+4)+(4j+2)) = (2)(8j+6) = 16j+12$

$(4j+3)^2 - (4j+1)^2 = ((4j+3)-(4j+1))((4j+3)+(4j+1)) = (2)(8j+4) = 16j+8$

So, $P_j = (16j+12) + (16j+8) = 32j+20$.

The series $S_n$ consists of $n$ such blocks (since the sum goes up to $4n$, there are $n$ blocks of 4 terms, for $j=0, 1, \dots, n-1$).

$S_n = \sum_{j=0}^{n-1} P_j = \sum_{j=0}^{n-1} (32j+20)$

$S_n = 32 \sum_{j=0}^{n-1} j + \sum_{j=0}^{n-1} 20$

Using the sum of the first $(n-1)$ integers, $\sum_{j=0}^{n-1} j = \frac{(n-1)n}{2}$:

$S_n = 32 \frac{n(n-1)}{2} + 20n$

$S_n = 16n(n-1) + 20n$

$S_n = 16n^2 - 16n + 20n$

$S_n = 16n^2 + 4n$

$S_n = 4n(4n+1)$

Now we test the given values to see which ones $S_n$ can take. For $S_n$ to be a possible value, $n$ must be a positive integer.

1. For $S_n = 1056$:

$4n(4n+1) = 1056$

$16n^2 + 4n - 1056 = 0$

Divide by 4: $4n^2 + n - 264 = 0$

Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$n = \frac{-1 \pm \sqrt{1^2 - 4(4)(-264)}}{2(4)} = \frac{-1 \pm \sqrt{1 + 4224}}{8} = \frac{-1 \pm \sqrt{4225}}{8}$

Since $65^2 = 4225$:

$n = \frac{-1 \pm 65}{8}$

Since $n$ must be positive, $n = \frac{-1 + 65}{8} = \frac{64}{8} = 8$.

Since $n=8$ is a positive integer, 1056 is a possible value.

2. For $S_n = 1088$:

$4n(4n+1) = 1088$

$16n^2 + 4n - 1088 = 0$

Divide by 4: $4n^2 + n - 272 = 0$

$n = \frac{-1 \pm \sqrt{1^2 - 4(4)(-272)}}{2(4)} = \frac{-1 \pm \sqrt{1 + 4352}}{8} = \frac{-1 \pm \sqrt{4353}}{8}$

Since 4353 ends in 3, it is not a perfect square. Thus, $n$ is not an integer, and 1088 is not a possible value.

3. For $S_n = 1120$:

$4n(4n+1) = 1120$

$16n^2 + 4n - 1120 = 0$

Divide by 4: $4n^2 + n - 280 = 0$

$n = \frac{-1 \pm \sqrt{1^2 - 4(4)(-280)}}{2(4)} = \frac{-1 \pm \sqrt{1 + 4480}}{8} = \frac{-1 \pm \sqrt{4481}}{8}$

Since 4481 is not a perfect square ($60^2=3600, 70^2=4900$, $61^2=3721, 69^2=4761$), $n$ is not an integer. Thus, 1120 is not a possible value.

4. For $S_n = 1332$:

$4n(4n+1) = 1332$

$16n^2 + 4n - 1332 = 0$

Divide by 4: $4n^2 + n - 333 = 0$

$n = \frac{-1 \pm \sqrt{1^2 - 4(4)(-333)}}{2(4)} = \frac{-1 \pm \sqrt{1 + 5328}}{8} = \frac{-1 \pm \sqrt{5329}}{8}$

Since $73^2 = 5329$:

$n = \frac{-1 \pm 73}{8}$

Since $n$ must be positive, $n = \frac{-1 + 73}{8} = \frac{72}{8} = 9$.

Since $n=9$ is a positive integer, 1332 is a possible value.

Therefore, $S_n$ can take values 1056 and 1332.