Let the first term of a series be T_1=6 and its r^ term T_r=3T_r-1+6^r,r=2,3, ............

Question

Let the first term of a series be ${T}_{1}=6$ and its ${r}^{}$ term ${T}_{r}=3{T}_{r-1}+{6}^{r},r=2,3$,
............ $n$. 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 \_\_\_\_\_\_\_\_\_\_\_.

Answer: 6

Accepted Answer

$\begin{gathered} {T}_{r}=3{T}_{r-1}+{6}^{r} \ \Rightarrow \ {T}_{r}=3{T}_{r-1} \ \Rightarrow x=3 \end{gathered}$

$	herefore {T}_{r}=a{.3}^{r}$

Solving for particular part

$\begin{gathered} {T}_{r}=b{.6}^{r} \ b{.6}^{r}=3b{6}^{r-1}+{6}^{r} \ \Rightarrow 6b=3b+6 \ \Rightarrow 3b=6 \ \Rightarrow b=2 \ {T}_{r}={a}^{n}+{a}^{p} \ {T}_{r}=a{3}^{br}+{2.6}^{r}\, \ {T}_{r}=3{T}_{r-1}+{6}^{r} \end{gathered}$

Putting $r=2$

${T}_{2}=18+36=54\,$

Using equation (i) and (ii)

$\begin{gathered} 54=9a+72\Rightarrow -18=9a\Rightarrow a=-2 \ 	herefore {T}_{r}=2\cdot {6}^{r}-2\cdot {3}^{r}=2({6}^{r}-{3}^{r}) \ \sum _{r=1}^{n}{T}_{r}=2\sum {6}^{r}-2\sum {3}^{r} \ =2\cdot 6\frac{({6}^{n}-1)}{5}-2\cdot 3\frac{({3}^{n}-1)}{2} \ =\frac{3}{5}(4\cdot {6}^{n}-5\cdot {3}^{n}+1) \ 	herefore {n}^{2}-12n+39=3 \ {n}^{2}-12n+36=0 \ (n-6{)}^{2}=0 \ 	herefore n=6 \end{gathered}$

Solutions · 1