Let the first term of a series be T_1=6 and i | vedclass

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(C) Given $T_1=6$ and $T_r=3T_{r-1}+6^r$ for $r \ge 2$.Dividing by $3^r$,we get $\frac{T_r}{3^r} = \frac{T_{r-1}}{3^{r-1}} + 2^r$.Let $a_r = \frac{T_r

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$6$

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(C) Given $T_1=6$ and $T_r=3T_{r-1}+6^r$ for $r \ge 2$.Dividing by $3^r$,we get $\frac{T_r}{3^r} = \frac{T_{r-1}}{3^{r-1}} + 2^r$.Let $a_r = \frac{T_r}{3^r}$. Then $a_r = a_{r-1} + 2^r$ with $a_1 = \frac{T_1}{3} = 2$.Summing from $r=2$ to $n$,we get $a_n = a_1 + \sum_{k=2}^n 2^k = 2 + (2^2 + 2^3 + \ldots + 2^n) = 2 + \frac{4(2^{n-1}-1)}{2-1} = 2 + 2^{n+1} - 4 = 2^{n+1}-2$.Thus,$T_n = 3^n(2^{n+1}-2) = 2 \cdot 6^n - 2 \cdot 3^n$.The sum $S_n = \sum_{r=1}^n T_r = 2 \sum_{r=1}^n 6^r - 2 \sum_{r=1}^n 3^r$.$S_n = 2 \left[ \frac{6(6^n-1)}{6-1} \right] - 2 \left[ \frac{3(3^n-1)}{3-1} \right] = \frac{12}{5}(6^n-1) - 3(3^n-1) = \frac{12 \cdot 6^n - 12 - 15 \cdot 3^n + 15}{5} = \frac{1}{5}(12 \cdot 6^n - 15 \cdot 3^n + 3) = \frac{3}{5}(4 \cdot 6^n - 5 \cdot 3^n + 1)$.Comparing this with the given sum $\frac{1}{5}(n^2-12n+39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$,we have $n^2-12n+39 = 3$.$n^2-12n+36 = 0 \implies (n-6)^2 = 0 \implies n=6$.

Let the first term of a series be $T_1=6$ and its $r^{\text{th}}$ term $T_r=3T_{r-1}+6^r$ for $r=2, 3, \ldots, n$. If the sum of the first $n$ terms of this series is $\frac{1}{5}(n^2-12n+39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$,then $n$ is equal to:

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