Let 729,81,9,1, be a sequence and P_n denote the product of the first n terms of this sequ…

Question

Let $729,81,9,1,\ldots$ be a sequence and ${P}_{n}$ denote the product of the first $n$ terms of this sequence.

If $2\sum _{n=1}^{40}{({P}_{n})}^{\frac{1}{n}}=\frac{{3}^{\alpha }-1}{{3}^{\beta }}$ and $gcd(\alpha ,\beta )=1$, then

$\alpha +\beta $ is equal to

Accepted Answer

$729,81,9,1,\ldots$ is a sequence in G.P. Can be written as ${9}^{3},{9}^{2},{9}^{1},{9}^{0},\frac{1}{9},{(\frac{1}{9})}^{2},\ldots$

${P}_{n}$ is product of first $n$ term

$\begin{gathered} {P}_{n}=({9}^{3})({9}^{2})({9}^{1})\ldots \ {P}_{n}={9}^{3+2+1+0+(-1)+(-2)\ldots } \end{gathered}$

Clearly power of 9 are in A.P. with first term $=3$

common difference $=-1$

So, sum of first $n$ terms of this A.P. $=\frac{n}{2}[2	imes 3+(n-1)(-1)]$

$=\frac{n}{2}[6-n+1]=\frac{n}{2}(7-n)$

So, ${P}_{n}=(9{)}^{\frac{n}{2}(7-n)}$

It is given that

$\begin{gathered} 2\sum _{n=1}^{40}{({P}_{n})}^{1/n}=\frac{{3}^{\alpha }-1}{{3}^{\beta }} \ \ \sum _{n=1}^{40}{({P}_{n})}^{1/n}=\sum _{n=1}^{40}(9{)}^{\frac{n}{2}(7-n)	imes \frac{1}{n}}=\sum _{n=1}^{40}{9}^{\frac{7-n}{2}} \ \sum _{n=1}^{40}{({9}^{1/2})}^{7-n}=\sum _{n=1}^{40}(3{)}^{7-n}={3}^{7}\sum _{n=1}^{40}{(\frac{1}{3})}^{n} \ ={3}^{7}[\frac{1}{3}+{(\frac{1}{3})}^{2}+{(\frac{1}{3})}^{3}\cdots {(\frac{1}{3})}^{40}] \ ={3}^{7}[=\frac{1}{3}=\frac{1}{3}] \ ={3}^{7}	imes \frac{1}{3}[\frac{1-{(\frac{1}{3})}^{40}}{1-\frac{1}{3}}] \end{gathered}$

$\begin{gathered} \sum _{n=1}^{40}{({P}_{n})}^{1/n}={3}^{6}(\frac{\frac{{3}^{40}-1}{{3}^{40}}}{\frac{2}{3}})=\frac{{3}^{40}-1}{2\cdot {3}^{33}} \ \, 2\sum _{n=1}^{40}{({P}_{n})}^{1/n}=\frac{{3}^{40}-1}{{3}^{33}}=\frac{{3}^{\alpha }-1}{{3}^{\beta }} \ \alpha =40,\beta =33 \ (40,33)=1 \ \alpha +\beta =40+33=73 \end{gathered}$

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