Question:

Step 1: Identify the recurrence. Since $\alpha, \beta, \gamma$ are roots of $x^3 - 4x + 1 = 0$, by Newton's identity: $$T_n = 4T_{n-2} - T_{n-3}$$ (using $p=0$, $q=-4$, $r=-1$ from $x^3+px^2+qx+r=0$) Step 2: Compute initial values. $T_0 = 3$, $T_1 = \alpha+\beta+\gamma = 0$ (sum of roots), $T_2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 0 - 2(-4) = 8$, $T_3 = 3\alpha\beta\gamma = 3(-1) = -3$ (using $\alpha\beta\gamma = -1$, $\alpha\beta+\beta\gamma+\gamma\alpha = -4$). Step 3: Check option 1: $T_6 - 4T_4 = 3$. Using recurrence: $T_n - 4T_{n-2} = -T_{n-3}$, so $T_6 - 4T_4 = -T_3 = -(-3) = 3$. TRUE. Step 4: Check option 2: $T_{96} + 2T_{99} + T_{102} = 16T_{98}$. From recurrence $T_{n} = 4T_{n-2} - T_{n-3}$: $T_{102} = 4T_{100} - T_{99}$ $T_{99} = 4T_{97} - T_{96}$ Also $T_{100} = 4T_{98} - T_{97}$. $T_{96} + 2T_{99} + T_{102} = T_{96} + 2T_{99} + 4T_{100} - T_{99} = T_{96} + T_{99} + 4T_{100}$ $= T_{96} + (4T_{97}-T_{96}) + 4(4T_{98}-T_{97}) = 4T_{97} + 16T_{98} - 4T_{97} = 16T_{98}$. TRUE. Step 5: Check option 3: $T_{96}+2T_{99}+T_{102} = 16T_{100}$. From Step 4, this equals $16T_{98} \neq 16T_{100}$ in general. FALSE. Step 6: Check option 4: $[\alpha]+[\beta]+[\gamma] = -2$. Analyze roots of $f(x)=x^3-4x+1=0$. $f(0)=1>0$, $f(-1)=4>0$, $f(-2)=-1<0$, $f(2)=1>0$, $f(1)=-2<0$. Roots: one in $(-2,-1)$, one in $(0,1)$, one in $(1,2)$. So $[\alpha]=-2$, $[\beta]=0$, $[\gamma]=1$. Sum $= -2+0+1 = -1$... Re-check: $f(-3)=-27+12+1=-14<0$, $f(-2)=-8+8+1=1>0$, so root in $(-3,-2)$: $[\alpha]=-3$. $f(0)=1>0$, $f(1)=-2<0$: root in $(0,1)$: $[\beta]=0$. $f(1)=-2<0$, $f(2)=1>0$: root in $(1,2)$: $[\gamma]=1$. Sum $=-3+0+1=-2$. TRUE.