MHB Finding the Value of $S_n$: A Summation Problem

[Saitama]

Let

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

Then $S_n$ can take the value(s)

A)1056

B)1088

C)1120

D)1332

 

[chisigma]

Let

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

Then $S_n$ can take the value(s)

A)1056

B)1088

C)1120

D)1332

[sp]Writing the first terms You obtain...

$\displaystyle S_{n} = -1 -4 + 9 + 16 - 25 - 36 + 49 + 64 - 81 - 100 + 121 + 144 +... = 8 + 12 + 24 + 28 + 40 + 44 + ... =$

$\displaystyle = 20\ n + 32\ \frac{n\ (n - 1)}{2} = 16\ n^{2} + 4\ n\ (1)$

... and for n=8 is $S_{8} = 1056$ and n=9 is $S_{9}= 1332$...[/sp]

Kind regards

$\chi$ $\sigma$

 

[magneto1]

Spoiler

\begin{align}
S_n &= \sum_{k=1}^{4n} (-1)^{\frac{k(k+1)}{2}} k^2 \\
&= \sum_{k=1}^n [ (-1)^{\frac{(4k-3)(4k-2)}{2}} (4k-3)^2 +
(-1)^{\frac{(4k-2)(4k-1)}{2}} (4k-2)^2 +
+(-1)^{\frac{(4k-1)(4k)}{2}} (4k-1)^2 +
(-1)^{\frac{(4k)(4+1)}{2}} (4k)^2 ] \\
&= \sum_{k=1}^n [ -(4k-3)^2 +
-(4k-2)^2
+(4k-1)^2
+(4k)^2 ] \\
&= \sum_{k=1}^n [ -(16k^2 - 24k + 9) - (16k^2 - 16k + 4) + (16k^2 - 8k + 1) + 16k^2] \\
&= \sum_{k=1}^n [ 32k - 12] \\
&= 16n(n+1)-12n
\end{align}

Enumerate through a few $n$, we find that (a) and (d) are possible values (for $n=8$ and $n=9$ respectively).

Instead of enumerating through $n$, we can also just spot check each answer.
We can actually do better to find all $X$ that this will work for: $16n^2 + 4n = X$. So the quadratic equation
would be $16n^2 - 4n - X = 0$. The discriminant $\Delta = 16+64X$ has to be a perfect square.

Note that when $X = 1056$, $\Delta = 260^2$ and when $X = 1332$, then $\Delta = 292^2$.

 

[Saitama]

[sp]Writing the first terms You obtain...

$\displaystyle S_{n} = -1 -4 + 9 + 16 - 25 - 36 + 49 + 64 - 81 - 100 + 121 + 144 +... = 8 + 12 + 24 + 28 + 40 + 44 + ... =$

$\displaystyle = 20\ n + 32\ (n - 1) = 52 n - 32\ (1)$

... so that $S_{n} - 32$ must divide 52 and that happens for 1332 because $\displaystyle \frac{1300}{52} = 25$...[/sp]

Kind regards

$\chi$ $\sigma$

Spoiler

\begin{align}
S_n &= \sum_{k=1}^{4n} (-1)^{\frac{k(k+1)}{2}} k^2 \\
&= \sum_{k=1}^n [ (-1)^{\frac{(4k-3)(4k-2)}{2}} (4k-3)^2 +
(-1)^{\frac{(4k-2)(4k-1)}{2}} (4k-2)^2 +
+(-1)^{\frac{(4k-1)(4k)}{2}} (4k-1)^2 +
(-1)^{\frac{(4k)(4+1)}{2}} (4k)^2 ] \\
&= \sum_{k=1}^n [ -(4k-3)^2 +
-(4k-2)^2
+(4k-1)^2
+(4k)^2 ] \\
&= \sum_{k=1}^n [ -(16k^2 - 24k + 9) - (16k^2 - 16k + 4) + (16k^2 - 8k + 1) + 16k^2] \\
&= \sum_{k=1}^n [ 32k - 12] \\
&= 16n(n+1)-12n
\end{align}

Enumerate through a few $n$, we find that (a) and (d) are possible values (for $n=8$ and $n=9$ respectively).

Instead of enumerating through $n$, we can also just spot check each answer.
We can actually do better to find all $X$ that this will work for: $16n^2 + 4n = X$. So the quadratic equation
would be $16n^2 - 4n - X = 0$. The discriminant $\Delta = 16+64X$ has to be a perfect square.

Note that when $X = 1056$, $\Delta = 260^2$ and when $X = 1332$, then $\Delta = 292^2$.

Thank you both for your participation, your answers are correct! :)

chisigma, you missed one of the answers. :p