Let $A = [\begin{matrix} 1 & \frac{3}{2} \\ 1 & 2 \end{matrix}], B = [\begin{matrix} 4 & - 3 \\ - 2 & 2 \end{matrix}] a n d C_{r} = [\begin{matrix} r {.3}^{r} & 2^{r} \\ 0 & (r - 1) 3^{r} \end{matrix}]$ be given matrices. If $\sum_{r = 1}^{50} t r ({(A B)}^{r} C_{r}) = 3 + a {.3}^{b}$ where $t r (A)$ denotes trace of matrix $A,$ then the value of $\frac{a + b}{25}$ is [Where $a$ and $b$ are relatively prime].

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Detailed Solution

We have $A B = [\begin{matrix} 1 & \frac{3}{2} \\ 1 & 2 \end{matrix}] [\begin{matrix} 4 & - 3 \\ - 2 & 2 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I$
$(A B)^{1} C_{1} = C_{1}, (A B)^{2} C_{2} = C_{2}$ and so on.
$t r (C_{r}) = r {.3}^{r} + (r - 1) {.3}^{r} = (2 r - 1) {.3}^{r} \sum_{r = 1}^{50} t r ({(A B)}^{r} C_{r}) = t r ({(A B)}^{1} C_{1}) + t r ({(A B)}^{2} C_{2}) + ..... + t r ({(A B)}^{50} C_{50}) = S (L e t) S = t r (C_{1}) + t r (C_{2}) + ..... + t r (C_{50}) S = {1.3}^{1} + {3.3}^{2} + {5.3}^{3} + ..... + {99.3}^{50} 3 S = {1.3}^{2} + {3.3}^{3} + ..... + {97.3}^{50} + {99.3}^{51} - 2 S = 1.3 + {2.3}^{2} + {3.3}^{3} + ..... + {2.3}^{50} - {99.3}^{51} = - 3 + 2.3 + {2.3}^{2} + ...... + {2.3}^{50} - {99.3}^{51} = - 3 + 2. \frac{3. (3^{50} - 1)}{3 - 1} - {99.3}^{51} = - 3 + 3^{51} - 3 - {99.3}^{51} = - 6 - {98.3}^{51} \Rightarrow S = 3 + {49.3}^{51} ∴ a + b = 100$