Let 729, 81, 9, 1,$729, 81, 9, 1, \dots$ be a sequence and Pₙ$P_n$ denote the product of the first n$n$ terms of this sequence. If 2Σn=1⁴⁰(Pₙ)(1)/(n) = (3^α - 1)/(3^β)$2\sum_{n=1}^{40}(P_n)^{\frac{1}{n}} = \frac{3^\alpha - 1}{3^\beta}$ and (α, β) = 1$\gcd(\alpha, \beta) = 1$, then α + β$\alpha + \beta$ is equal to

Core Logic

Let the given sum be S$S$. Exclude the first term and identify the progression in the remaining terms:

Wait, rewrite the terms carefully: Term 2: 10 · 13²⁵$\frac{10 \cdot 1}{3^{25}}$ Term 3: 10 · 23²⁴ = 103²⁵ · (2 · 3)$\frac{10 \cdot 2}{3^{24}} = \frac{10}{3^{25}} \cdot (2 \cdot 3)$ Term 4: 10 · 2²3²³ = 103²⁵ · (2² · 3²) = 103²⁵ · (6²)$\frac{10 \cdot 2^2}{3^{23}} = \frac{10}{3^{25}} \cdot (2^2 \cdot 3^2) = \frac{10}{3^{25}} \cdot (6^2)$ Term n$n$: 10 · 2²⁴3¹ = 103²⁵ · 2²⁴ · 3²⁴ = 103²⁵ · (6)²⁴$\frac{10 \cdot 2^{24}}{3^1} = \frac{10}{3^{25}} \cdot 2^{24} \cdot 3^{24} = \frac{10}{3^{25}} \cdot (6)^{24}$

Execution

The bracketed terms form a geometric progression with common ratio r = 6$r = 6$, and total terms = 25.

Sum of G.P. = 6²⁵ - 16 - 1 = 6²⁵ - 15$= \frac{6^{25} - 1}{6 - 1} = \frac{6^{25} - 1}{5}$.

Substitute this back:

Pattern Recognition

Factoring out the lowest common denominator scaling factor transforms chaotic-looking polynomial denominators into a straightforward integer geometric series block.

Chapter Mix

Class 11 Maths: Sequence and Series

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Core Logic

Factor the denominator: (r)/(r⁴ + r² + 1) = (r)/((r² - r + 1)(r² + r + 1))$\frac{r}{r^4 + r^2 + 1} = \frac{r}{(r^2 - r + 1)(r^2 + r + 1)}$ Decompose using partial fractions (Method of Differences):

Execution

Let the sum be S$S$.

Expand the sum to observe telescoping cancellation: r=1: ((1)/(1) - (1)/(3))$r=1: \left(\frac{1}{1} - \frac{1}{3}\right)$ r=2: ((1)/(3) - (1)/(7))$r=2: \left(\frac{1}{3} - \frac{1}{7}\right)$ r=3: ((1)/(7) - (1)/(13))$r=3: \left(\frac{1}{7} - \frac{1}{13}\right)$ ... r=25: ((1)/(601) - (1)/(651))$r=25: \left(\frac{1}{601} - \frac{1}{651}\right)$

Summing all terms leaves only the first and last parts:

Since gcd(325, 651) = 1$\text{gcd}(325, 651) = 1$, we have p = 325$p = 325$ and q = 651$q = 651$. p + q = 325 + 651 = 976$p + q = 325 + 651 = 976$.

Pattern Recognition

The expression r⁴ + r² + 1$r^4 + r^2 + 1$ is the canonical telescoping denominator. Instantly split it into difference of squares (r²+1)² - r²$(r^2+1)^2 - r^2$ to unleash the cancellation chain.

Chapter Mix

Class 11 Maths: Sequence and Series

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Core Logic

Let the A.P. have n$n$ terms (where n$n$ is even). The terms are divided into n/2$n/2$ odd-indexed terms and n/2$n/2$ even-indexed terms.

Step 1: Set up the even and odd sums

Sum of even terms:

Sum of odd terms:

Subtracting equation (2) from (1):

Since there are n/2$n/2$ such pairs, and the difference of adjacent terms is the common difference d$d$:

Step 2: Solve for n and d

We are given that the last term exceeds the first by (21)/(2)$\frac{21}{2}$:

n d - d = 10.5$n d - d = 10.5$

Substitute nd = 12$nd = 12$ from (3):

Using this in (3):

Step 3: Solve for the first term

The sum of the odd terms is:

Thus, the terms are:

The terms that are integers are 3, 6, 9, 12$3, 6, 9, 12$. The total number of integer terms is 4.

Pattern Recognition

Sum of even terms minus sum of odd terms in any A.P. with an even number of terms n$n$ is always equal to (n)/(2) d$\frac{n}{2} d$. This is an extremely useful relation to remember.

Chapter Mix

Class 11 Mathematics: Sequences and Series

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Core Logic

This is a telescoping series sum. We expand the denominator using Sophie Germain's algebraic identity to write the general term as a difference of two consecutive rational expressions.

Step 1: Write down the general term and factor

The general term Tᵣ$T_r$ of the series is:

Using the factorization of 1+4r⁴$1+4r^4$:

Notice that the numerator 4r$4r$ is the exact difference of the two quadratic factors:

Step 2: Split the fraction into telescoping terms

Rewrite the general term Tᵣ$T_r$:

Step 3: Expand the sum and solve

Sum the first 10 terms:

• For r = 1$r = 1$: T₁ = (1)/(1) - (1)/(5)$T_1 = \frac{1}{1} - \frac{1}{5}$
• For r = 2$r = 2$: T₂ = (1)/(5) - (1)/(13)$T_2 = \frac{1}{5} - \frac{1}{13}$
• ...
• For r = 10$r = 10$: T₁₀ = (1)/(181) - (1)/(221)$T_{10} = \frac{1}{181} - \frac{1}{221}$

All intermediate terms cancel out:

Since 220$220$ and 221$221$ are coprime (their greatest common divisor is 1):

Pattern Recognition

Sophie Germain identity: The expansion 4r⁴ + 1 = (2r² - 2r + 1)(2r² + 2r + 1)$4r^4 + 1 = (2r^2 - 2r + 1)(2r^2 + 2r + 1)$ is highly common in telescoping series. Spotting this factorization collapses the sum instantly.

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Class 11 Mathematics: Sequences and Series

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Sum of an Arithmetic Progression:

Core Logic

Express the odd terms summation in terms of a₁$a_1$ and d$d$, establish their linear dependency, and then solve for n$n$ where total sum vanishes.

Step 1: Simplify the Given Summation

The summation represents the sum of 12$12$ terms: a₁ + a₃ + a₅ + + a₂₃$a_1 + a_3 + a_5 + \dots + a_{23}$. This is an A.P. with initial term a₁$a_1$ and common difference 2d$2d$.

Step 2: Relate initial term to common difference

Multiply through by 5$5$ to eliminate the fraction:

Step 3: Solve for n

Set the general sum of n$n$ terms to 0$0$:

Substitute a₁ = -5d$a_1 = -5d$:

Since a₁ ≠ 0$a_1 \neq 0$, d ≠ 0$d \neq 0$, which yields:

n = 11$n = 11$

Pattern Recognition

The condition a₁ = -5d$a_1 = -5d$ means the sequence starts positive/negative and counts down symmetrically. A sum of n$n$ terms equals zero when the middle term or balanced pairs completely wipe each other out, pointing directly to n = 2(5) + 1 = 11$n = 2(5) + 1 = 11$.

Chapter Mix

Class 11 Mathematics: Sequences and Series