Question

If $T_r \in N; r \in N$,

$\sum_{r=1}^{n} T_r = \frac{1}{5}(n^2 - 12n + 39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$,

$T_r = 3T_{r-1} + 6^r \forall 2 \leq r \leq n$, $T_1 = 6$,

then the value of $n$, is equal to

Answer 1

6

Answer 2

The problem provides a recurrence relation for a sequence $T_r$ and an expression for its sum $S_n = \sum_{r=1}^{n} T_r$. We need to find the value of $n$.

Step 1: Solve the recurrence relation for $T_r$. The given recurrence relation is $T_r = 3T_{r-1} + 6^r$ for $2 \leq r \leq n$, with $T_1 = 6$. To solve this, divide by $3^r$: $\frac{T_r}{3^r} = \frac{3T_{r-1}}{3^r} + \frac{6^r}{3^r}$ $\frac{T_r}{3^r} = \frac{T_{r-1}}{3^{r-1}} + 2^r$ Let $U_r = \frac{T_r}{3^r}$. The recurrence relation transforms into: $U_r = U_{r-1} + 2^r \quad \text{for } r \geq 2$ First, find $U_1$: $U_1 = \frac{T_1}{3^1} = \frac{6}{3} = 2$ Now, express $U_r$ as a sum: $U_r = U_1 + \sum_{k=2}^{r} 2^k$ $U_r = 2 + (2^2 + 2^3 + \dots + 2^r)$ The sum in the parenthesis is a geometric progression with first term $a = 2^2 = 4$, common ratio $R = 2$, and number of terms $(r-1)$. The sum of this geometric progression is $\frac{a(R^{r-1}-1)}{R-1} = \frac{4(2^{r-1}-1)}{2-1} = 4(2^{r-1}-1) = 2^2 \cdot 2^{r-1} - 4 = 2^{r+1} - 4$. Substitute this back into the expression for $U_r$: $U_r = 2 + (2^{r+1} - 4) = 2^{r+1} - 2$ Now, substitute back $U_r = \frac{T_r}{3^r}$: $\frac{T_r}{3^r} = 2^{r+1} - 2$ $T_r = 3^r (2^{r+1} - 2) = 3^r \cdot 2 \cdot (2^r - 1)$ $T_r = 2 \cdot 3^r \cdot 2^r - 2 \cdot 3^r$ $T_r = 2 \cdot 6^r - 2 \cdot 3^r$ Let's verify for $T_1$: $T_1 = 2 \cdot 6^1 - 2 \cdot 3^1 = 12 - 6 = 6$, which matches the given $T_1$.

Step 2: Calculate the sum $S_n = \sum_{r=1}^{n} T_r$. $S_n = \sum_{r=1}^{n} (2 \cdot 6^r - 2 \cdot 3^r)$ $S_n = 2 \sum_{r=1}^{n} 6^r - 2 \sum_{r=1}^{n} 3^r$ Both sums are geometric progressions. For $\sum_{r=1}^{n} 6^r$: first term $a_1=6$, common ratio $R_1=6$. Sum is $\frac{6(6^n-1)}{6-1} = \frac{6}{5}(6^n-1)$. For $\sum_{r=1}^{n} 3^r$: first term $a_2=3$, common ratio $R_2=3$. Sum is $\frac{3(3^n-1)}{3-1} = \frac{3}{2}(3^n-1)$. Substitute these sums back into the expression for $S_n$: $S_n = 2 \cdot \frac{6}{5}(6^n-1) - 2 \cdot \frac{3}{2}(3^n-1)$ $S_n = \frac{12}{5}(6^n-1) - 3(3^n-1)$ $S_n = \frac{12}{5} 6^n - \frac{12}{5} - 3 \cdot 3^n + 3$ $S_n = \frac{12}{5} 6^n - 3 \cdot 3^n + \left(3 - \frac{12}{5}\right)$ $S_n = \frac{12}{5} 6^n - 3 \cdot 3^n + \frac{15-12}{5}$ $S_n = \frac{12}{5} 6^n - \frac{15}{5} 3^n + \frac{3}{5}$ Factor out $\frac{1}{5}$: $S_n = \frac{1}{5} (12 \cdot 6^n - 15 \cdot 3^n + 3)$ Further factor out 3 from the terms inside the parenthesis: $S_n = \frac{1}{5} \cdot 3 \cdot (4 \cdot 6^n - 5 \cdot 3^n + 1)$

Step 3: Compare the derived $S_n$ with the given $S_n$. The given sum is: $S_n = \frac{1}{5}(n^2 - 12n + 39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$ Comparing this with our derived sum: $\frac{1}{5} \cdot 3 \cdot (4 \cdot 6^n - 5 \cdot 3^n + 1) = \frac{1}{5}(n^2 - 12n + 39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$ For this equality to hold, the coefficients of $(4 \cdot 6^n - 5 \cdot 3^n + 1)$ must be equal (assuming $(4 \cdot 6^n - 5 \cdot 3^n + 1) \neq 0$, which it is not for $n \in N$): $3 = n^2 - 12n + 39$

Step 4: Solve the quadratic equation for $n$. $n^2 - 12n + 39 - 3 = 0$ $n^2 - 12n + 36 = 0$ This is a perfect square trinomial: $(n-6)^2 = 0$ $n-6 = 0$ $n = 6$