Aspire Faculty ID #18039 · Topic: JEE Main 2026 (24 January Morning Shift) · Just now
JEE Main 2026 (24 January Morning Shift)

Let $729, 81, 9, 1, \ldots$ be a sequence and $P_n$ denote the product of the first $n$ terms of this sequence. If $2\sum_{n=1}^{40}(P_n)=\dfrac{3^\alpha-1}{3^\beta}$ and $\gcd(\alpha,\beta)=1$, then $\alpha+\beta$ is equal to:

Solution

$P_n = 729 \cdot 81 \cdot 9 \cdot \ldots ;(\text{n terms})$

$= 3^6 \cdot 3^4 \cdot 3^2 \cdot \ldots$

$P_n = 3^{6+4+2+\cdots} = 3^{n(7-n)}$

$\Rightarrow \sum_{n=1}^{40} (P_n) = 3^6 + 3^5 + \cdots + (40\ \text{terms})$

$= 3^6\left[\dfrac{1-(\frac{1}{3})^{40}}{1-\frac{1}{3}}\right]$

$= \dfrac{3^6(3^{40}-1)}{2\cdot 3^{40}}$

$= \dfrac{3^{40}-1}{2\cdot 3^{33}}$

Comparing

$\alpha = 40,; \beta = 33$

$\Rightarrow \alpha+\beta = 73$
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