Perhaps you call it Newton's sum

Since $\alpha$ and $\beta$ are roots of $x^2+x-1=0$ . we have:

$\begin{cases} \alpha^2+\alpha-1=0 \\ \beta^2+\beta-1=0 \end{cases} \Rightarrow \begin{cases} \alpha^2+\alpha^1-\alpha^0=0 \\ \beta^2+\beta^1-\beta^0=0 \end{cases}$

$\Rightarrow (\alpha^2+\beta^2) + (\alpha^1 + \beta^1 ) - ( \alpha ^0 + \beta^0) = 0 \quad \Rightarrow V_2=-V_1+V_0$

Similarly,

$\begin{cases} \alpha^3+\alpha^2-\alpha^1=0 \\ \beta^3+\beta^2-\beta^1=0 \end{cases} \Rightarrow V_3=-V_2+V_1$

$\begin{cases} \alpha^4+\alpha^3-\alpha^2=0 \\ \beta^4+\beta^3-\beta^2=0 \end{cases} \Rightarrow V_4=-V_3+V_2 \quad \Rightarrow V_n = - V_{n-1} + V_{n-2}$

$V_0 = \alpha^0 + \beta^0 = 2$

$V_1 = \alpha+ \beta = -1$

$V_2 = - V_1 + V_0 = 1 + 2 = 3$

$V_3 = - V_2 + V_1 = -3 -1 = -4$

$V_4 = - V_3 + V_2 = 4 + 3 = 7$

$V_5 = - V_4 + V_3 = - 7 - 4 = -11$

$V_6 = - V_5 + V_4 = 11 + 7 = 18$

$V_7 = - V_6 + V_5 = -18 -11 = \boxed{-29}$

$V_n+V_{n-3}=[\alpha^n+\beta^n]+[ \alpha^{n-3} +\beta^{n-3} ]$

$\quad \quad \quad \quad =\alpha^{n-3}(\alpha^3+1) + \beta^{n-3}(\beta^3+1)$

And

$\begin{cases} \alpha^3+1=(\alpha^2+\alpha -1)(\alpha-1)+2.\alpha = 2. \alpha \\ \beta^3+1=(\beta^2+\beta-1)(\beta-1)+2.\beta=2.\beta \end{cases}$

Therefore

$V_n+V_{n-3}=\alpha^{n-3}(2.\alpha) + \beta^{n-3}(2.\beta)$

$\implies \large \boxed{V_n+ V_{n-3}=2.V_{n-2}}$

$\implies V_7=2.V_5-V_4=2.(2.V_3-V_2)-(2V_2-V_1)=4.V_3-4.V_2+V_1$

$\implies V_7=4.(2V_1-V_0)-4.V_2+V_1=9.V_1-4.V_0-4.V_2$

$\begin{cases} V_0=\alpha^0+\beta^0=2. \\ V_1=\alpha+\beta=-1 \\ V_2=\alpha^2+\beta^2=3 \end{cases}$

So

$V_7=\boxed{-29}$

$From\ the\ question\ we\ have:\\ \alpha+\beta=(-1).......................................(1)\\ \alpha*\beta=(-1)........................................(2)\\ .Squaring\ the\ first\ equation\ gives:\\ \alpha^{2}+\beta^{2}+2*\alpha*\beta=1.\\Substituting\ the\ value\ of \alpha*\beta gives:\\ \alpha^{2}+\beta^{2}=3. Again\ squaring\ the\ above\ equation\ and\ substituting\ the\ value\\\ of\ \alpha*\beta\ gives\ that:\alpha^{4}+\beta^{4}=7(3).Now,let\ us\ find\ out\ the\\\ value\ of\ \alpha^{3}+\beta^{3}.\\ We\ cube\ the\ first\ equation\ to\ get:\\ \alpha^{3}+\beta^{3}+3*\alpha*\beta(\alpha+\beta)=(-1)(4).\\ Substituting\ the\ values\ of\ \alpha+\beta\ and\ \alpha*\beta\ gives:\\ \alpha^{3}+\beta^{3}=(-4).Multiplying\ equations\ (3)\ and\ (4)\ we\ get:\\ \alpha^{7}+\beta^{7}+\alpha^{3}*\beta^{3}(\alpha+\beta)=(-28).Substitute\ the\\\ values\ of\ \alpha+\beta\ and\ \alpha*\beta\ and\ get\ the\ answer.$

$x^{7}=(x^{2}+x-1)(x^{5}-x^{4}+2x^{3}-3x^{2}+5x-8)+13x-8$

For $x=\alpha$ and $x=\beta$ ,

$x^{7}=13x-8$ since $x^{2}+x-1=0$

$\begin{aligned} &\alpha^{7}+\beta^{7} \\ &=(13 \alpha -8)+(13 \beta-8) \\ &=13 (\alpha + \beta) -16 \\ &=13 (-1)-16 & \small \color{#3D99F6}{ \text{ by Vieta} } \\ &= \boxed {-29} \end{aligned}$

$\begin{aligned}{x}^2&=1-x\\{x}^3&=x-{x}^2\\&=2x-1\\{x}^6&=4{x}^2-4x+1\\&=5-8x\\{x}^7&=5x-8{x}^2\\&=13x-8\\V_7&=13(\alpha+\beta)-16\\&\huge\color{#3D99F6}{=\boxed{-29}}\end{aligned}$

$On \; solving\; the\; equation, \; we \; get \newline \alpha = \frac{-1+\sqrt{5}}{2} and \; \beta = \frac{-1-\sqrt{5}}{2} \newline \: \newline$ $We\;have \; x^{2} = 1-x$ $\newline$ $\therefore$ $x^{4}=1-2x-x^{2}$ $\newline$ $=1-2x+1-x$ $\newline$ $=2-3x$ $\newline$ $Similarly, \; x^{8}=4-12x+9x^{2}$ $\newline$ $=4-12x+9(1-x)$ $\newline$ $=13-21x$ $\newline$ $\therefore x^{7} = \frac{13}{x}-21$ $\newline \: \newline$ $Since \; x^{2} = 1-x,\; x=\frac{1}{x} -1$ $\newline$ $Hence, \; \frac{1}{x} = x +1$ $\newline \: \newline$ $\therefore x^{7} = 13(x+1) -21 = 13x-8$ $\newline \; \newline$ $\therefore V_7 = \alpha^{7} + \beta^{7} \newline =13\alpha - 8 +13\beta -8 \newline = 13(\alpha +\beta) -16$ $\newline$ $=13(\frac{-1+\sqrt{5}-1-\sqrt{5}}{2}) -16$ $\newline$ $=\boxed{-29}$