Let 729, 81, 9, 1, ... be a sequence and Pn denote the product of the first n terms of this sequence

Q: Why is the exponent an arithmetic progression? I thought the sequence was geometric.

That is a brilliant observation! The sequence of terms $729, 81, 9, 1$ is indeed a geometric progression with a common ratio of $1/9$. However, when we express these terms as powers of $3$, we get $3^6, 3^4, 3^2, 3^0$. If you look only at the exponents—$6, 4, 2, 0$—you will see that they decrease by a constant value of $2$. That is the definition of an arithmetic progression. We are essentially mapping a geometric sequence into an arithmetic one by using logarithms/exponents.

Q: How do I know when to use the GP sum formula?

Whenever you see a series where each term is a constant multiple of the previous one, it is a geometric progression (GP). In our expanded series $3^6 + 3^5 + \dots + 3^{-33}$, each term is obtained by multiplying the previous one by $1/3$. Since the ratio is constant ($r = 1/3$), you can safely use the GP sum formula: $S = a \frac{1-r^N}{1-r}$.

Q: Why did we need to take the LCM at the very end?

The problem asks us to match our result to the specific form $\frac{3^\alpha - 1}{3^\beta}$. Our initial calculation gave us $3^7 - 3^{-33}$. To combine these into a single fraction, we need a common denominator. By writing $3^{-33}$ as $\frac{1}{3^{33}}$, we can easily find the common denominator, which allows us to identify $\alpha$ and $\beta$ by direct comparison.

Question

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

Options:

(A) 74

(B) 76

(C) 73

(D) 75

Sigma Percentile · Accepted Answer

### The Beauty of Hidden Patterns

Welcome, fellow traveler in the world of mathematics. Today, we are going to dismantle a problem that, at first glance, might seem like a chaotic mess of exponents and products.

As we peel back the layers, you will see that it is actually a beautifully orchestrated sequence. Let us embark on this journey together.

### Phase 1: Decoding the Sequence

We are given the sequence $729, 81, 9, 1, \dots$. When you see numbers like these, your intuition should immediately scream, "Powers of three!"

Let us verify this:
$729 = 3^6$
$81 = 3^4$
$9 = 3^2$
$1 = 3^0$

Look at the exponents: $6, 4, 2, 0$. They form an arithmetic progression (AP) where the first term $a = 6$ and the common difference $d = -2$.

The general term for the exponent is $6 + (k-1)(-2)$, which simplifies to $8 - 2k$. Thus, the $k$-th term of our sequence is $T_k = 3^{8-2k}$.

### Phase 2: The Product $P_n$

The problem defines $P_n$ as the product of the first $n$ terms. This means $P_n = T_1 \cdot T_2 \cdot \dots \cdot T_n$.

Using our general term, this is:
$$P_n = \prod_{k=1}^n 3^{8-2k}$$

Recall the fundamental law of exponents: $a^x \cdot a^y = a^{x+y}$. When we multiply these terms, we are essentially summing the exponents.

So, $P_n = 3^{\sum_{k=1}^n (8-2k)}$. Let us calculate that sum of exponents, $S_{exp} = \sum_{k=1}^n (8-2k)$.

We can split this into two parts: $\sum_{k=1}^n 8 - 2 \sum_{k=1}^n k$. The sum of the constant $8$ taken $n$ times is $8n$, and the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.

Putting it together:
$$S_{exp} = 8n - 2 \cdot \frac{n(n+1)}{2} = 8n - (n^2 + n) = 7n - n^2$$

Thus, our product is $P_n = 3^{7n - n^2}$.

### Phase 3: The $n$-th Root

The problem asks us to work with $(P_n)^{1/n}$. This looks intimidating, but watch how it collapses:

$$(P_n)^{1/n} = (3^{7n - n^2})^{1/n} = 3^{\frac{7n - n^2}{n}} = 3^{7-n}$$

The complexity vanishes, leaving us with a simple expression: $3^{7-n}$.

### Phase 4: The Summation

We are now tasked with evaluating $2 \sum_{n=1}^{40} (P_n)^{1/n}$, which is $2 \sum_{n=1}^{40} 3^{7-n}$.

Let us write out the first few terms to see the structure:
For $n=1$, we get $3^6$.
For $n=2$, we get $3^5$.
For $n=40$, we get $3^{7-40} = 3^{-33}$.

This is a geometric progression with first term $a = 3^6$, common ratio $r = 1/3$, and $N = 40$ terms. Using the sum formula $S_{GP} = a \frac{1-r^N}{1-r}$, we get:

$$Sum = 3^6 \cdot \frac{1 - (1/3)^{40}}{1 - 1/3} = 3^6 \cdot \frac{1 - 3^{-40}}{2/3} = \frac{3^7}{2} (1 - 3^{-40})$$

### Phase 5: The Final Match

Don't forget the factor of $2$ outside the summation! Multiplying our sum by $2$, we get:

$$2 \cdot \frac{3^7}{2} (1 - 3^{-40}) = 3^7 (1 - 3^{-40}) = 3^7 - 3^{-33}$$

To match the form $\frac{3^\alpha - 1}{3^\beta}$, we write $3^{-33}$ as $\frac{1}{3^{33}}$ and take the common denominator:

$$\frac{3^7 \cdot 3^{33} - 1}{3^{33}} = \frac{3^{40} - 1}{3^{33}}$$

By direct comparison, $\alpha = 40$ and $\beta = 33$. Since $\gcd(40, 33) = 1$, our values are correct.

The final answer is **$\alpha + \beta = 40 + 33 = 73$**.

Animated Solution for Mathematics - Sequence and Series: Let $729, 81, 9, 1, \dots$ be a sequence and $P_{n}$ denote the product of the first $n$ terms of this sequence. If $2 \sum_{n = 1}^{40} (P_{n})^{1/ n} = \frac{3 ^{α} - 1}{3 ^{β}}$ and $g cd (α, β) = 1$ , then $α + β$ is equal to

Select Answer:

Visualized Solution

Analyzing the Sequence

Finding the General Term $T_{k}$

Calculating the Product $P_{n}$

Simplifying the Exponent Sum

Result for $P_{n}$ and $(P_{n})^{\frac{1}{n}}$

Setting up the Final Summation

Expanding the Series

Applying the GP Sum Formula

Calculating the Sum Value

Finding $2 S$ and Simplifying

Matching the Given Form

Identifying $α$ and $β$

The Sigma Insight: Geometric Progression (G.P.)

The Beauty of Hidden Patterns

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

The sum of first four terms of a geometric progression (G.P.) is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18}$ . If the product of first three terms of the G.P. is 1, and the third term is $α$ , then $2 α$ is

In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is $R - (a, b)$ , then $a^{2} + b^{2}$ is equal to ......... .

Let $a_{1}, a_{2}, a_{3}, \dots$ be a GP of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then $a_{1} a_{9} + a_{2} a_{4} a_{9} + a_{5} + a_{7}$ is equal to

Let $a_{1}, a_{2}, a_{3}, \dots$ be a G. P. of increasing positive numbers. If $a_{3} a_{5} = 729$ and $a_{2} + a_{4} = \frac{111}{4}$ , then $24 (a_{1} + a_{2} + a_{3})$ is equal to

In an increasing geometric progression of positive terms, the sum of the second and sixth terms is $\frac{70}{3}$ and the product of the third and fifth terms is 49. Then the sum of the $4^{th}, 6^{th}$ and $8^{th}$ terms is equal to :

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

Let the first term $a$ and the common ratio $r$ of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to

Let $a$ and $b$ be two distinct positive real numbers. Let $1 1^{t h}$ term of a GP, whose first term is $a$ and third term is $b$ , is equal to $p^{t h}$ term of another GP, whose first term is $a$ and fifth term is $b$ . Then $p$ is equal to

Let $a_{1}, a_{2}, a_{3}, \dots$ be G.P. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4} = 64$ and $a_{1} + a_{3} + a_{5} = \frac{813}{7}$ . Then $a_{3} + a_{5} + a_{7}$ is equal to :