If (1)/(1⁴) + (1)/(2⁴) + (1)/(3⁴) + ∞ = (π⁴)/(90)$\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \dots \infty = \frac{\pi^4}{90}$, and (1)/(1⁴) + (1)/(3⁴) + (1)/(5⁴) + ∞ = α$$\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \dots \infty = \alpha$$ (1)/(2⁴) + (1)/(4⁴) + (1)/(6⁴) + ∞ = β$$\frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \dots \infty = \beta$$ then (α)/(β)$\frac{\alpha}{\beta}$ is equal to

### Related Formula n-th term of a GP: Tₙ = a rⁿ⁻¹$$n\text{-th term of a GP}: T_n = a r^{n-1}$$ ### Core Logic Let the terms of the GP be a, ar, ar², ar³,$a, ar, ar^2, ar^3, \dots$ Given: a₂ · a₃ · a₄ = 64$a_2 \cdot a_3 \cdot a_4 = 64$ (ar) · (ar²) · (ar³) = 64$$(ar) \cdot (ar^2) \cdot (ar^3) = 64$$ a³ r⁶ = 64 ⇒ (ar²)³ = 64 ⇒ ar² = 4$$a^3 r^6 = 64 \Rightarrow (ar^2)^3 = 64 \Rightarrow ar^2 = 4$$ Since it is a GP of increasing positive terms, r > 1$r > 1$ and a > 0$a > 0$. ### Step 1: Utilize the sum condition Given: a₁ + a₃ + a₅ = (813)/(7)$a_1 + a_3 + a_5 = \frac{813}{7}$ a + ar² + ar⁴ = (813)/(7)$$a + ar^2 + ar^4 = \frac{813}{7}$$ Extract a$a$ from ar² = 4 ⇒ a = (4)/(r²)$ar^2 = 4 \Rightarrow a = \frac{4}{r^2}$. Substitute this in: (4)/(r²) + 4 + 4r² = (813)/(7)$$\frac{4}{r^2} + 4 + 4r^2 = \frac{813}{7}$$ 4((1)/(r²) + 1 + r²) = (813)/(7)$$4\left(\frac{1}{r^2} + 1 + r^2\right) = \frac{813}{7}$$ Let r² = t$r^2 = t$: 4((1)/(t) + 1 + t) = (813)/(7)$$4\left(\frac{1}{t} + 1 + t\right) = \frac{813}{7}$$ Wait, there is a much faster method by just scaling the required expression. ### Step 2: Calculate the required expression We need a₃ + a₅ + a₇ = ar² + ar⁴ + ar⁶$a_3 + a_5 + a_7 = ar^2 + ar^4 + ar^6$. Notice that ar² + ar⁴ + ar⁶ = r² (a + ar² + ar⁴)$ar^2 + ar^4 + ar^6 = r^2 (a + ar^2 + ar^4)$. So, required sum = r² ((813)/(7))$r^2 \left(\frac{813}{7}\right)$. To find r²$r^2$, we solve the quadratic in t = r²$t = r^2$: 4((t² + t + 1)/(t)) = (813)/(7)$$4\left(\frac{t^2 + t + 1}{t}\right) = \frac{813}{7}$$ 28t² + 28t + 28 = 813t$$28t^2 + 28t + 28 = 813t$$ 28t² - 785t + 28 = 0$$28t^2 - 785t + 28 = 0$$ The roots are t = 28$t = 28$ and t = (1)/(28)$t = \frac{1}{28}$. Since the GP is increasing, r > 1 ⇒ r² = 28$r > 1 \Rightarrow r^2 = 28$. ### Step 3: Final evaluation Alternatively, expand directly: ar² (1 + r² + r⁴) = 4 (1 + 28 + (28)²) = 4(1 + 28 + 784) = 4(813) = 3252.$$ar^2 (1 + r^2 + r^4) = 4 (1 + 28 + (28)^2) = 4(1 + 28 + 784) = 4(813) = 3252.$$ (Note: r² × (813)/(7) = 28 × (813)/(7) = 4 × 813 = 3252$r^2 \times \frac{813}{7} = 28 \times \frac{813}{7} = 4 \times 813 = 3252$) ### Pattern Recognition In GP questions demanding a sum shifted by a fixed index (like a₁+a₃+a₅$a_1+a_3+a_5$ to a₃+a₅+a₇$a_3+a_5+a_7$), immediately look to factor out the common ratio multiplier r^k$r^k$. Here it's a simple scaling by r²$r^2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series Class 10 Maths: Quadratic Equations

### Related Formula Partial fraction decomposition for telescopic summing: (2n² - (n+1)² + 1 )/( ) structures directly cancel in series expansions.$$\frac{2n^2 - (n+1)^2 + 1 \dots}{\dots} \text{ structures directly cancel in series expansions.}$$ ### Core Logic Given recurrence: aₙ₊₁ - (1)/(2)aₙ = (n² - 2n - 1)/(n²(n+1)²)$$a_{n+1} - \frac{1}{2}a_n = \frac{n^2 - 2n - 1}{n^2(n+1)^2}$$ Rewrite the numerator to split the fraction: n² - 2n - 1 = 2n² - (n² + 2n + 1) = 2n² - (n+1)²$$n^2 - 2n - 1 = 2n^2 - (n^2 + 2n + 1) = 2n^2 - (n+1)^2$$ aₙ₊₁ - (1)/(2)aₙ = (2n² - (n+1)²)/(n²(n+1)²) = (2)/((n+1)²) - (1)/(n²)$$a_{n+1} - \frac{1}{2}a_n = \frac{2n^2 - (n+1)^2}{n^2(n+1)^2} = \frac{2}{(n+1)^2} - \frac{1}{n^2}$$ ### Step 1: Telescope generation Multiply both sides by appropriate powers of 2 to create a cancelling chain: For n=1$n=1$: a₂ - (1)/(2)a₁ = (2)/(2²) - (1)/(1²)$a_2 - \frac{1}{2}a_1 = \frac{2}{2^2} - \frac{1}{1^2}$ For n=2$n=2$: multiply by 2 ⇒ 2[a₃ - (1)/(2)a₂ = (2)/(3²) - (1)/(2²)] ⇒ 2a₃ - a₂ = (2 × 2)/(3²) - (2)/(2²)$\Rightarrow 2\left[a_3 - \frac{1}{2}a_2 = \frac{2}{3^2} - \frac{1}{2^2}\right] \Rightarrow 2a_3 - a_2 = \frac{2 \times 2}{3^2} - \frac{2}{2^2}$ Wait, let's look at a cleaner telescopic scaling: aₙ₊₁ - (2)/((n+1)²) = (1)/(2) (aₙ - (2)/(n²))$a_{n+1} - \frac{2}{(n+1)^2} = \frac{1}{2} \left(a_n - \frac{2}{n^2}\right)$. ### Step 2: Identify Geometric Progression Let Vₙ = aₙ - (2)/(n²)$V_n = a_n - \frac{2}{n^2}$. The recurrence gives Vₙ₊₁ = (1)/(2) Vₙ$V_{n+1} = \frac{1}{2} V_n$. This proves Vₙ$V_n$ is a geometric progression with common ratio r = 1/2$r = 1/2$. First term V₁ = a₁ - (2)/(1²) = 1 - 2 = -1$V_1 = a_1 - \frac{2}{1^2} = 1 - 2 = -1$. ### Step 3: Infinite Summation We need | Σn=1∞ ( aₙ - (2)/(n²) ) | = | Σn=1∞ Vₙ |$\left| \sum_{n=1}^{\infty} \left( a_n - \frac{2}{n^2} \right) \right| = \left| \sum_{n=1}^{\infty} V_n \right|$. Since Vₙ$V_n$ is an infinite GP: S∞ = (V₁)/(1 - r) = (-1)/(1 - 1/2) = (-1)/(1/2) = -2$$S_{\infty} = \frac{V_1}{1 - r} = \frac{-1}{1 - 1/2} = \frac{-1}{1/2} = -2$$ Taking absolute value: |-2| = 2$|-2| = 2$ ### Pattern Recognition When dealing with rational fraction recurrences Aₙ₊₁ - k Aₙ = f(n) - k f(n-1)$A_{n+1} - k A_n = f(n) - k f(n-1)$, immediately substitute Vₙ = Aₙ - f(n)$V_n = A_n - f(n)$. This substitution instantly isolates a classical Geometric Progression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

### Related Formula Sum of G.P. Sₙ = (a(rⁿ - 1))/(r - 1) (for r > 1)$$\text{Sum of G.P. } S_n = \frac{a(r^n - 1)}{r - 1} \text{ (for } r > 1)$$ ### Core Logic The given sequence is a G.P. with ratio 1√(2)$\frac{1}{\sqrt{2}}$. (a₂/2)/(a₁) = 1√(2) a₂ = a₁ √(2)$$\frac{a_2/2}{a_1} = \frac{1}{\sqrt{2}} \implies a_2 = a_1 \sqrt{2}$$ (a₃/2²)/(a₂/2) = 1√(2) (a₃)/(2 a₂) = 1√(2) a₃ = a₂ √(2) = a₁ (√(2))²$$\frac{a_3/2^2}{a_2/2} = \frac{1}{\sqrt{2}} \implies \frac{a_3}{2 a_2} = \frac{1}{\sqrt{2}} \implies a_3 = a_2 \sqrt{2} = a_1 (\sqrt{2})^2$$ Thus, a₁, a₂, a₃, , a₁₀$a_1, a_2, a_3, \dots, a_{10}$ forms a standard G.P. with first term a₁$a_1$ and common ratio R = √(2)$R = \sqrt{2}$. ### Step 1: Calculate the Sum Sum of this new sequence is S₁₀ = 62$S_{10} = 62$. S₁₀ = a₁ ( (√(2))¹⁰ - 1 )√(2) - 1 = 62$$S_{10} = \frac{a_1 \left( (\sqrt{2})^{10} - 1 \right)}{\sqrt{2} - 1} = 62$$ Since (√(2))¹⁰ = 2⁵ = 32$(\sqrt{2})^{10} = 2^5 = 32$: 62 = a₁ (32 - 1)√(2) - 1$$62 = \frac{a_1 (32 - 1)}{\sqrt{2} - 1}$$ 62 = 31 a₁√(2) - 1$$62 = \frac{31 a_1}{\sqrt{2} - 1}$$ 2 = a₁√(2) - 1$$2 = \frac{a_1}{\sqrt{2} - 1}$$ a₁ = 2(√(2) - 1)$$a_1 = 2(\sqrt{2} - 1)$$ ### Pattern Recognition If a sequence bₙ = aₙkⁿ⁻¹$b_n = \frac{a_n}{k^{n-1}}$ is a G.P. with ratio r$r$, then the base sequence aₙ$a_n$ is inherently a G.P. with ratio R = kr$R = kr$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequence and Series

### Related Formula Sₙ = (n)/(2)[2a + (n-1)d]$$S_n = \frac{n}{2}[2a + (n-1)d]$$ ### Core Logic Given the sum of first 4 terms is S₄ = 6$S_4 = 6$: (4)/(2)(2a + 3d) = 6 2a + 3d = 3 (1)$$\frac{4}{2}(2a + 3d) = 6 \implies 2a + 3d = 3 \quad \dots (1)$$ Given the sum of first 6 terms is S₆ = 4$S_6 = 4$: (6)/(2)(2a + 5d) = 4 2a + 5d = (4)/(3) (2)$$\frac{6}{2}(2a + 5d) = 4 \implies 2a + 5d = \frac{4}{3} \quad \dots (2)$$ ### Step 1: Finding Parameters a and d Subtract equation (1) from equation (2): (2a + 5d) - (2a + 3d) = (4)/(3) - 3$$(2a + 5d) - (2a + 3d) = \frac{4}{3} - 3$$ 2d = -(5)/(3) d = -(5)/(6)$$2d = -\frac{5}{3} \implies d = -\frac{5}{6}$$ Substitute d$d$ into equation (1): 2a + 3(-(5)/(6)) = 3$$2a + 3\left(-\frac{5}{6}\right) = 3$$ 2a - (5)/(2) = 3$$2a - \frac{5}{2} = 3$$ 2a = 3 + (5)/(2) = (11)/(2) a = (11)/(4)$$2a = 3 + \frac{5}{2} = \frac{11}{2} \implies a = \frac{11}{4}$$ ### Step 2: Calculating Sum of 12 terms S₁₂ = (12)/(2)[2a + 11d]$$S_{12} = \frac{12}{2}[2a + 11d]$$ S₁₂ = 6 [2((11)/(4)) + 11(-(5)/(6))]$$S_{12} = 6 \left[2\left(\frac{11}{4}\right) + 11\left(-\frac{5}{6}\right)\right]$$ S₁₂ = 6 [ (11)/(2) - (55)/(6) ] = 6 [ (33 - 55)/(6) ]$$S_{12} = 6 \left[ \frac{11}{2} - \frac{55}{6} \right] = 6 \left[ \frac{33 - 55}{6} \right]$$ S₁₂ = 33 - 55 = -22$$S_{12} = 33 - 55 = -22$$ ### Pattern Recognition Standard two-variable linear equations format strictly from sum identities. Fractions easily resolve by distributing the external multiplication constant n/2$n/2$ directly into the bracket. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

### Related Formula Exponential functional equation: f(x+y) = f(x)f(y) f(x) = a^x$f(x+y) = f(x)f(y) \implies f(x) = a^x$. Geometric progression sum formula: Sₙ = (a(rⁿ - 1))/(r - 1)$S_n = \frac{a(r^n - 1)}{r - 1}$. ### Core Logic Since f(1) = 7$f(1) = 7$, f(x) = 7^x$f(x) = 7^x$. For g(x+y) = g(xy)$g(x+y) = g(xy)$, set y = 1 g(x+1) = g(x)$y = 1 \implies g(x+1) = g(x)$. Since g(1) = 1$g(1) = 1$, we have g(1) = g(2) = = g(n) = 1$g(1) = g(2) = \dots = g(n) = 1$. ### Step 1: Solve Summation Σx=1ⁿ (7^x)/(1) = 19607 7 ( (7ⁿ - 1)/(7 - 1) ) = 19607$$\sum_{x=1}^{n} \frac{7^x}{1} = 19607 \implies 7 \left( \frac{7^n - 1}{7 - 1} \right) = 19607$$ (7)/(6) (7ⁿ - 1) = 19607 7ⁿ - 1 = 16806 7ⁿ = 16807$$\frac{7}{6} (7^n - 1) = 19607 \implies 7^n - 1 = 16806 \implies 7^n = 16807$$ Since 7⁵ = 16807$7^5 = 16807$, n = 5$n = 5$. ### Pattern Recognition Recognize f(x)=7^x$f(x)=7^x$ and constant function g(x)=1$g(x)=1$ from given functional equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series Class 11 Maths: Functions and Graphs