This question was designed based on this other question here.
Evaluate $$\sum_{k=1}^{2018}(-1)^{^{\binom {k+2}3}}k$$ and explain why it would be a perfect square. Generalize.
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Hypergeometricx
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the result is giveb by$$1018081$$ – Dr. Sonnhard Graubner May 25 '17 at 08:28
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The sequence of the signs is $-1,+1,+1,+1$ . I hope this helps. :-) – user90369 May 25 '17 at 14:21
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We have that every $4$th term starting from the first is odd so numbers of the form $4k+1$ have a negative sign so we have that the sum is equal to. $$\sum_{k=1}^{4t+2}(-1)^{k+2\choose 3}k=\sum_{k=1}^{4t+2}k-2\sum_{k=0}^{t}(4k+1)=(2t+1)(4t+3)-4t(t+1)-2(t+1)=(2t+1)(4t+3)-2(t+1)(2t+1)=(2t+1)(4t+3-2t-2)=(2t+1)^2$$
kingW3
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Sign change
Note that $\displaystyle\binom {k+2}3=\frac{(k+2)(k+1)k}6$.
As $2$ is a factor of the denominator $6$ , $\displaystyle \binom {k+2}3$ is odd if the numerator has only one multiple of $2$ (but not of higher powers of $2$) in the denominator, and even otherwise.
if $k$ is odd then, $k+2$ is also odd and only $k+1$ is even, and would be either
(i) a multiple of $4$, in which case $\binom {k+2}3$ is even,
or
(ii) a multiple of $2$ but not of $4$, in which case $\binom {k+2}3$ is odd (as $2$ would cancel with $6$ leaving $3$ in the denominator and no more $2$'s left in the numerator. This occurs every $4r+1$ term.
if $k$ is even then $k+2$ is also even, hence $\binom {k+2}3$ is even.
Hence the signs for $(-1)^{\binom {k+2}3}$ for $k=1,2,3,\dots$ are $\color{red}-+++\color{red}-+++\color{red}-+++\color{red}-+++\cdots$
Summation
Note that
$$\begin{align} \sum_{k=1}^{4m+2}(-1)^{\binom {k+2}3}k &=-1+2+\sum_{r=1}^m (4r-1)+4r-(4r+1)+(4r+2)\\ &=1+\sum_{r=1}^m 8r\\ &=1+4m(m+1)\\ &=(2m+1)^2 \end{align}$$ Putting $m=504$ (note that $4\times 504+2=2018$) gives $$\sum_{k=1}^{2018}(-1)^{\binom{k+2}3}k=\color{red}{1009^2}$$
Special Note
It is interesting to note that $\displaystyle \sum_{k=1}^{4m+3}(-1)^{\binom {k+2}3}k$ is also a perfect square. This is shown below.
$$\begin{align} \sum_{k=1}^{4m+3}(-1)^{\binom {k+2}3}k &=-1+2+3+\sum_{r=1}^m 4r-(4r+1)+(4r+2)+(4r+3)\\ &=4+4\sum_{r=1}^m (2r+1)\\ &=4+4\left[(m+1)^2-1\right] &&\scriptsize\text{as }(2r+1)=(r+1)^2-r^2\\ &=4(m+1)^2\\ &=\left[2(m+1)\right]^2\\ &=(2m+1)^2 \end{align}$$
Hypergeometricx
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