MathFun4Kids

Solution: If $c=0$, based on equation $(2)$ and $(3)$, we have $a^2=1$ and $b^2=1$. Therefore, $a^2+b^2=2$, which is in conflict with equation $(1)$. Therefore $c\ne 0$.

Because $c\ne 0$, based on equation $(1)$ and $(3)$, we have:

$$a=\dfrac{b^2-c^2}{c}\tag{4}$$

Based on equation $(1)$ and $(2)$, we have:

$$a^2=c^2+\sqrt{3}bc\tag{5}$$

Replacing $a$ in equation $(5)$ with equation $(3)$, we have:

$$\Big(\dfrac{b^2-c^2}{c}\Big)^2=c^2+\sqrt{3}bc\tag{6}$$

Simpifying equation $(6)$, we have: $$b(b^3-2bc^2-\sqrt{3}c^3)=0\tag{7}$$

If $b=0$, we have $a^2=1$, $c^2=1$, $c^2+a^2+ca=1$, which leads solutions of $(a, b, c)$ as $(-1, 0, 1)$ and $(1, 0, -1)$.

If $(b^3-2bc^2-\sqrt{3}c^3=0$, as $c\ne 0$, we have: $$\Big(\dfrac{b}{c}\Big)^3-2\Big(\dfrac{b}{c}\Big)-\sqrt{3}=0\tag{8}$$

The rational solution for $\dfrac{b}{c}$ in equation $(8)$ is $\sqrt{3}$. Therefore $$b=\sqrt{3}c$$

Applying the above result in equation $(2)$, we have $$c=\pm\dfrac{\sqrt{7}}{7}$$

which also results in $$b=\pm\dfrac{\sqrt{21}}{7}$$ and $$a=\pm\dfrac{2\sqrt{7}}{7}$$

Combining the all cases, there are 4 real solutions for $(a, b, c)$:

$(-1,0,1)$, $(1,0,-1)$, $\Big(\dfrac{2\sqrt{7}}{7},\dfrac{\sqrt{21}}{7},\dfrac{\sqrt{7}}{7}\Big)$, and $\Big(-\dfrac{2\sqrt{7}}{7},-\dfrac{\sqrt{21}}{7},-\dfrac{\sqrt{7}}{7}\Big)$

Solution: If $n=1$, then

\begin{flalign*} \sum\limits_{i=1}^{n}(-1)^{i+1}\cdot i\cdot\binom{n-1}{i-1} &= (-1)^{1+1}\cdot 1\cdot\binom{1-1}{1-1}& \\ &=1& \end{flalign*}

If $n=2$, then

\begin{flalign*} \sum\limits_{i=1}^{n}(-1)^{i+1}\cdot i\cdot\binom{n-1}{i-1} &=(-1)^{1+1}\cdot 1\cdot\binom{2-1}{1-1}+(-1)^{2+1}\cdot 2\cdot\binom{2-1}{2-1}& \\ &=1-2& \\ &=-1& \end{flalign*}

If $n\ge 3$, then

\begin{flalign*} \sum\limits_{i=1}^{n}(-1)^{i+1}\cdot i\cdot\binom{n-1}{i-1} &= \sum\limits_{i=0}^{n-1}(-1)^{i+2}\cdot (i+1)\cdot\binom{n-1}{i} &\\ &=\sum\limits_{i=0}^{n-1}(-1)^i\cdot i\cdot \binom{n-1}{i}+\sum\limits_{i=0}^{n-1}(-1)^i\cdot\binom{n-1}{i}& \\ &=\sum\limits_{i=1}^{n-1}(-1)^i\cdot i\cdot \binom{n-1}{i}+(1-1)^{n-1} &\\ &=\sum\limits_{i=1}^{n-1}(-1)^i\cdot i\cdot \dfrac{(n-1)!}{i!\cdot ((n-1)-i)!} + 0 &\\ &=(n-1)\cdot\sum\limits_{i=1}^{n-1}(-1)^i\cdot \dfrac{(n-2)!}{(i-1)!\cdot ((n-2)-(i-1))!} &\\ &=(n-1)\cdot(-1)\cdot\sum\limits_{i=1}^{n-1}(-1)^{i-1}\cdot \binom{n-2}{i-1} &\\ &=(1-n)\cdot\sum\limits_{i=0}^{n-2}(-1)^i\cdot \binom{n-2}{i} &\\ &=(1-n)\cdot(1-1)^{n-2} &\\ &=0 & \end{flalign*}

Therefore, we have

\begin{flalign*} \sum\limits_{i=1}^{n}(-1)^{i+1}\cdot i\cdot\binom{n-1}{i-1}=& \left\{ \begin{array}{rl@{\qquad}l} 1 & n = 1 \\ -1 & n = 2 \\ 0 & n \ge 3 \\ \end{array} \right. & \\ \end{flalign*}