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JEE Mathematics 2026 Question with Solution

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

  • A

    7373

  • B

    7575

  • C

    7676

  • D

    7474

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The sequence is 729,81,9,1,…729, 81, 9, 1, \ldots and PnP_n is the product of the first nn terms.

Find: The value of α+β\alpha + \beta if

2∑n=140(Pn)1n=3α−13β.2 \sum_{n=1}^{40} (P_n)^{\frac{1}{n}} = \frac{3^{\alpha} - 1}{3^{\beta}}.

Write the terms as powers of 33:

729=36,81=34,9=32,1=30.729 = 3^6, \quad 81 = 3^4, \quad 9 = 3^2, \quad 1 = 3^0.

So the nnth term is

an=32(4−n).a_n = 3^{2(4-n)}.

Now find PnP_n:

Pn=∏k=1n32(4−k)=32∑k=1n(4−k).P_n = \prod_{k=1}^{n} 3^{2(4-k)} = 3^{2\sum_{k=1}^{n}(4-k)}.

Also,

∑k=1n(4−k)=4n−n(n+1)2.\sum_{k=1}^{n}(4-k) = 4n - \frac{n(n+1)}{2}.

Hence,

Pn=32(4n−n(n+1)2)=38n−n(n+1)=37n−n2.P_n = 3^{2\left(4n - \frac{n(n+1)}{2}\right)} = 3^{8n - n(n+1)} = 3^{7n - n^2}.

Therefore,

(Pn)1n=37n−n2n=37−n.(P_n)^{\frac{1}{n}} = 3^{\frac{7n-n^2}{n}} = 3^{7-n}.

So the required sum becomes

2∑n=14037−n.2\sum_{n=1}^{40} 3^{7-n}.

Let k=7−nk = 7-n. Then as nn goes from 11 to 4040, kk goes from 66 to −33-33. Thus,

2∑n=14037−n=2∑k=−3363k.2\sum_{n=1}^{40} 3^{7-n} = 2\sum_{k=-33}^{6} 3^k.

This is a geometric series. Using

∑k=−3363k=3−33∑r=0393r=3−33⋅340−13−1,\sum_{k=-33}^{6} 3^k = 3^{-33}\sum_{r=0}^{39} 3^r = 3^{-33}\cdot \frac{3^{40}-1}{3-1},

we get

2∑k=−3363k=2⋅3−33⋅340−12=340−1333.2\sum_{k=-33}^{6} 3^k = 2 \cdot 3^{-33} \cdot \frac{3^{40}-1}{2} = \frac{3^{40}-1}{3^{33}}.

Comparing with

3α−13β,\frac{3^{\alpha}-1}{3^{\beta}},

we obtain

α=40,β=33.\alpha = 40, \quad \beta = 33.

Hence,

α+β=73.\alpha + \beta = 73.

Therefore, the correct option is A.

The solution states option B, but the extracted working gives 340−1333\frac{3^{40}-1}{3^{33}} and hence α+β=73\alpha+\beta = 73. Therefore the worked solution supports option A, not B.

Geometric Series Expansion

After obtaining

(Pn)1n=37−n,(P_n)^{\frac{1}{n}} = 3^{7-n},

the sum is

2(36+35+⋯+3−33).2(3^6 + 3^5 + \cdots + 3^{-33}).

This can be written as

2(3−33+3−32+⋯+36).2\left(3^{-33} + 3^{-32} + \cdots + 3^6\right).

Here the first term is 3−333^{-33}, the common ratio is 33, and the number of terms is 4040. Therefore,

∑k=−3363k=3−33(340−1)3−1.\sum_{k=-33}^{6} 3^k = \frac{3^{-33}(3^{40}-1)}{3-1}.

Multiplying by 22 gives

2∑k=−3363k=2⋅3−33(340−1)2=340−1333.2\sum_{k=-33}^{6} 3^k = 2 \cdot \frac{3^{-33}(3^{40}-1)}{2} = \frac{3^{40}-1}{3^{33}}.

So again α=40\alpha = 40 and β=33\beta = 33, which gives α+β=73\alpha+\beta = 73.

Common mistakes

  • A common mistake is writing the general term incorrectly. The exponents go as 6,4,2,0,…6, 4, 2, 0, \ldots, so the correct form is an=32(4−n)a_n = 3^{2(4-n)}. If this exponent pattern is misread, every later step becomes incorrect.

  • Another mistake is evaluating ∑k=1n(4−k)\sum_{k=1}^{n}(4-k) incorrectly. It is 4n−n(n+1)24n - \frac{n(n+1)}{2}, not 4n−n4n-n. Use the sum of the first nn natural numbers carefully.

  • Students often mishandle the geometric series with negative exponents. Rewrite the series as 3−33∑r=0393r3^{-33}\sum_{r=0}^{39}3^r before applying the finite geometric series formula.

  • A frequent error is trusting the listed correct option even when the algebra shows otherwise. When the working yields 340−1333\frac{3^{40}-1}{3^{33}}, the correct conclusion is α+β=73\alpha+\beta=73, so the defensible answer is option A.

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