Let 729,81,9,1,… be a G.P., and let Pn denote the product of the first n terms of this sequence. If
\[\sum_{n=1}^{40} (P_n)^{1/n} = \frac{3^{\alpha}-1}{2\cdot 3^{\beta}}, \quad \gcd(\alpha,\beta)=1,\]
then α+β is equal to:
Solutions · 1
Official solution
The G.P. is 93,92,91,90,9−1,… (ratio 91), so the exponents of 9 form an A.P. with first term 3, common difference −1.
\[P_n = 9^{3+2+\cdots+(4-n)} = 9^{\frac{n}{2}(7-n)} \implies (P_n)^{1/n} = 9^{(7-n)/2} = 3^{7-n}.\]
\[\sum_{n=1}^{40} 3^{7-n} = 3^7\sum_{n=1}^{40}\left(\frac13\right)^n = 3^7\cdot\frac{\tfrac13\left(1-3^{-40}\right)}{1-\tfrac13} = \frac{3^7}{2}\left(1-3^{-40}\right) = \frac{3^{40}-1}{2\cdot 3^{33}}.\]
Comparing, α=40, β=33, gcd(40,33)=1, so α+β=73.