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JEE Main 202624 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2026 · 24 January · Morning Shift · Mathematics · Sequences and Series

Let 729,81,9,1, be a sequence and P_ n denote the product of the...

Let 729,81,9,1,…729,81,9,1, \ldots be a sequence and Pn\mathrm{P}_{n} denote the product of the first nn terms of this sequence.
If 2∑n=140(Pn)1n=3α−13β2 \sum_{n=1}^{40}\left(\mathrm{P}_{n}\right)^{\frac{1}{n}}=\frac{3^{\alpha}-1}{3^{\beta}} and gcd⁡(α,β)=1\operatorname{gcd}(\alpha, \beta)=1,
then α+β\alpha+\beta is equal to

Options

  1. A74
  2. B76
  3. C73
  4. D75

Correct answer

C. 73

Step-by-step solution

The sequence is 36,34,32,30,3−2,…3^6, 3^4, 3^2, 3^0, 3^{-2}, \ldots so an=38−2na_n = 3^{8-2n}.
Pn=∏k=1n38−2k=3∑k=1n(8−2k)=38n−n(n+1)=3n(7−n)P_n = \prod_{k=1}^n 3^{8-2k} = 3^{\sum_{k=1}^n(8-2k)} = 3^{8n - n(n+1)} = 3^{n(7-n)}
(Pn)1/n=37−n(P_n)^{1/n} = 3^{7-n}
2∑n=14037−n=2(36+35+⋯+3−33)=2⋅36⋅1−3−401−1/3=37−3−332\sum_{n=1}^{40} 3^{7-n} = 2(3^6+3^5+\cdots+3^{-33}) = 2 \cdot 3^6 \cdot \frac{1-3^{-40}}{1-1/3} = 3^7 - 3^{-33}
=340−1333= \frac{3^{40}-1}{3^{33}}
So α=40,  β=33\alpha = 40,\; \beta = 33.
Since gcd⁡(40,33)=1\gcd(40,33)=1, we get α+β=73\alpha+\beta = 73.
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