Isn't 100000 such a huge number (Part 11 ) ?

Newton's Identities says that if

$f(x) = {a}_{n}{x}^{n} + {a}_{n-1}{x}^{n-1} + ... + {a}_{1}x + {a}_{0}$ with roots ${r}_{1}, {r}_{2},... {r}_{n}$ and ${V}_{n} = {{r}_{1}}^{n} + {{r}_{2}}^{n} + .... + {{r}_{n}}^{n}$ , then

${V}_{n}{a}_{n} + {V}_{n-1}{a}_{n-1} + .... + {V}_{1}{a}_{1} + n{a}_{0} = 0$

Now, all one has to do is substitute, starting with ${V}_{1} + 1 = 0$ .

As we go further, $1$ will be added to the constant term and ${V}_{n+1}$ to the equation. To balance the equation, $1 - 1 = 0 \Rightarrow {V}_{n} = -1$

Hence, $\displaystyle \sum_{n=0}^{100000}{{(-1)}^{n}. {V}_{n}} = 124 -(-1) + (-1)... + (-1) - (-1) + (-1) = 124$

$T\left( x \right) =\cfrac { { x }^{ 125 }-1 }{ x-1 } =0,x\neq 1\\ \\ \Rightarrow x={ e }^{ \cfrac { ^{ 2m\pi } }{ 125 } }={ \alpha }^{ m }\quad for\quad m=1,2,3,...,124,{ a }_{ 1 }=\alpha \\ \\ \Rightarrow { V }_{ n }={ \alpha }^{ n }{ +\alpha }^{ 2n }{ +...+\alpha }^{ 124n }=\cfrac { { \left( { \alpha }^{ 125 } \right) }^{ n }-1 }{ { \alpha }^{ n }-1 } =0\quad only\quad if\quad { \alpha }^{ n }-1\neq 0\\ \\ if\quad { \alpha }^{ n }-1=0\Rightarrow { V }_{ n }=124\quad only\quad if\quad n\quad is\quad multiple\quad of\quad 125\\ \\ \Rightarrow { V }_{ n }=\left\{ \begin{matrix} 0\quad for\quad n\neq 125r\quad ,r=0,1,2,...,800 \\ 124\quad for\quad n=125r\quad ,r=0,1,2,...,800 \end{matrix} \right\} \\ \\ \Rightarrow S=\sum _{ n=0 }^{ 100000 }{ { \left( -1 \right) }^{ n } } { V }_{ n }=\sum _{ r=0 }^{ 800 }{ { \left( -1 \right) }^{ 125r } } \left( 124 \right) \\ \\ \quad \quad \quad =124\sum _{ r=0 }^{ 800 }{ { \left( -1 \right) }^{ r } } =124\left\{ 0+{ \left( -1 \right) }^{ 800 } \right\} \\ \\ \quad \quad =124$