Question:

Step 1: Given $T_{n+1} = T_n^3 + 2T_n$. So $T_{n+1} = T_n(T_n^2 + 2)$. Step 2: Evaluate the sum $\sum_{n=1}^{100}(T_n^3 + T_n + 1)$. Note $T_n^3 + 2T_n = T_{n+1}$, so $T_n^3 = T_{n+1} - 2T_n$. Thus $T_n^3 + T_n + 1 = T_{n+1} - 2T_n + T_n + 1 = T_{n+1} - T_n + 1$. Step 3: Sum: $\sum_{n=1}^{100}(T_{n+1} - T_n + 1) = (T_{101} - T_1) + 100 = T_{101} - 1 + 100 = T_{101} + 99$. So option (a) is correct. Step 4: Check option (c): $\prod_{n=1}^{100}(T_n^2 + 2)$. From $T_{n+1} = T_n(T_n^2 + 2)$, we get $T_n^2 + 2 = \frac{T_{n+1}}{T_n}$. So $\prod_{n=1}^{100}(T_n^2 + 2) = \prod_{n=1}^{100}\frac{T_{n+1}}{T_n} = \frac{T_{101}}{T_1} = T_{101}$ (telescoping), not $T_{100}$. So option (c) is incorrect. Step 5: Check option (d): $\prod_{n=1}^{100}\left(\frac{T_{n+1}}{T_n} - T_n^2\right)$. Since $\frac{T_{n+1}}{T_n} = T_n^2 + 2$, we get $\frac{T_{n+1}}{T_n} - T_n^2 = 2$ for each $n$. So $\prod_{n=1}^{100} 2 = 2^{100}$. Option (d) is correct.