The sum 1 + frac1+32! + frac1+3+53! + frac1+3+5+74! + dots$1 + \frac{1+3}{2!} + \frac{1+3+5}{3!} + \frac{1+3+5+7}{4!} + \dots$ is equal to

### Related Formula Sum of first n$n$ terms of an Arithmetic Progression: S_n = fracn2[2a + (n-1)d]$$S_n = \frac{n}{2}[2a + (n-1)d]$$ ### Core Logic Set up linear expressions for the given sums : S_40 = frac402[2a + 39d] = 1030 Rightarrow 2a + 39d = 51.5$$S_{40} = \frac{40}{2}[2a + 39d] = 1030 \Rightarrow 2a + 39d = 51.5$$ S_12 = frac122[2a + 11d] = 57 Rightarrow 2a + 11d = 9.5$$S_{12} = \frac{12}{2}[2a + 11d] = 57 \Rightarrow 2a + 11d = 9.5$$ ### Step 1: Solve for a$a$ and d$d$ Subtract the second equation from the first : (2a + 39d) - (2a + 11d) = 51.5 - 9.5$$(2a + 39d) - (2a + 11d) = 51.5 - 9.5$$ 28d = 42 Rightarrow d = frac4228 = frac32 = 1.5$$28d = 42 \Rightarrow d = \frac{42}{28} = \frac{3}{2} = 1.5$$ Substitute d = 1.5$d = 1.5$ back to find a$a$: 2a + 11(1.5) = 9.5 Rightarrow 2a + 16.5 = 9.5 Rightarrow 2a = -7 Rightarrow a = -3.5$$2a + 11(1.5) = 9.5 \Rightarrow 2a + 16.5 = 9.5 \Rightarrow 2a = -7 \Rightarrow a = -3.5$$ ### Step 2: Evaluate S_30 - S_10$S_{30} - S_{10}$ Write out the formula for the target subtraction : S_30 - S_10 = frac302[2a + 29d] - frac102[2a + 9d]$$S_{30} - S_{10} = \frac{30}{2}[2a + 29d] - \frac{10}{2}[2a + 9d]$$ = 15(2a + 29d) - 5(2a + 9d) = 30a + 435d - 10a - 45d = 20a + 390d$$= 15(2a + 29d) - 5(2a + 9d) = 30a + 435d - 10a - 45d = 20a + 390d$$ [cite: 3964, 3965] Substitute the values of a$a$ and d$d$ : = 20(-3.5) + 390(1.5) = -70 + 585 = 515$$= 20(-3.5) + 390(1.5) = -70 + 585 = 515$$ ### Pattern Recognition Notice that S_30 - S_10$S_{30} - S_{10}$ represents the \sum of terms from T_11$T_{11}$ to T_30$T_{30}$, which can also be formulated as 20 times A_20.5$20 \times A_{20.5}$, saving algebraic steps if calculated symmetrically. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula Sum of an infinite geometric progression: S_infty = fraca1-r quad textfor |r| < 1$$S_{\infty} = \frac{a}{1-r} \quad \text{for } |r| < 1$$ ### Core Logic The given expression is an infinite Arithmetico-Geometric Progression (AGP) : S = 5 + frac5+alpha7 + frac5+2alpha7^2 + frac5+3alpha7^3 + dots infty$$S = 5 + \frac{5+\alpha}{7} + \frac{5+2\alpha}{7^2} + \frac{5+3\alpha}{7^3} + \dots \infty$$ ### Step 1: Shift and Subtract Multiply the equation by the common ratio frac17$\frac{1}{7}$ and shift it by one position : frac17S = frac57 + frac5+alpha7^2 + frac5+2alpha7^3 + dots infty$$\frac{1}{7}S = \frac{5}{7} + \frac{5+\alpha}{7^2} + \frac{5+2\alpha}{7^3} + \dots \infty$$ Subtract this from the original equation: S - frac17S = 5 + left(frac5+alpha-57right) + left(frac5+2alpha-(5+alpha)7^2right) + dots$$S - \frac{1}{7}S = 5 + \left(\frac{5+\alpha-5}{7}\right) + \left(\frac{5+2\alpha-(5+\alpha)}{7^2}\right) + \dots$$ frac67S = 5 + fracalpha7 + fracalpha7^2 + fracalpha7^3 + dots$$\frac{6}{7}S = 5 + \frac{\alpha}{7} + \frac{\alpha}{7^2} + \frac{\alpha}{7^3} + \dots$$ ### Step 2: Sum the Infinite Geometric Series Apply the infinite GP formula to the terms involving alpha$\alpha$ : frac67S = 5 + fracalpha7left(frac11 - frac17right) = 5 + fracalpha7left(frac76right) = 5 + fracalpha6$$\frac{6}{7}S = 5 + \frac{\alpha}{7}\left(\frac{1}{1 - \frac{1}{7}}\right) = 5 + \frac{\alpha}{7}\left(\frac{7}{6}\right) = 5 + \frac{\alpha}{6}$$ Given that S = 7$S = 7$ : frac67(7) = 5 + fracalpha6 Rightarrow 6 = 5 + fracalpha6$$\frac{6}{7}(7) = 5 + \frac{\alpha}{6} \Rightarrow 6 = 5 + \frac{\alpha}{6}$$ 1 = fracalpha6 Rightarrow alpha = 6$$1 = \frac{\alpha}{6} \Rightarrow \alpha = 6$$ ### Pattern Recognition Standard trick for infinite AGPs: Multiply by the common ratio r$r$, shift, and subtract to condense the arithmetic progression component into a straightforward infinite geometric progression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula The general term for the provided series is: T_k = frac1k(k+1) = frac1k - frac1k+1$$T_k = \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$$ This sets up a standard telescoping summation sequence. ### Core Logic Express the sum S_2025$S_{2025}$ via telescoping fractions: S_2025 = sum_k=1^2025 left( frac1k - frac1k+1 right) = left(1 - frac12right) + left(frac12 - frac13right) + dots + left(frac12025 - frac12026right)$$S_{2025} = \sum_{k=1}^{2025} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{2025} - \frac{1}{2026}\right)$$ S_2025 = 1 - frac12026 = frac20252026$$S_{2025} = 1 - \frac{1}{2026} = \frac{2025}{2026}$$ ### Step 1: Compute the boundary expression value Substitute S_2025$S_{2025}$ into the expression value: sqrt2026 cdot S_2025 = sqrt2026 cdot frac20252026 = sqrt2025 = 45$$\sqrt{2026 \cdot S_{2025}} = \sqrt{2026 \cdot \frac{2025}{2026}} = \sqrt{2025} = 45$$ ### Step 2: Apply Arithmetic Progression Summation The sum of the first 6 terms of the A.P. with a = -p$a = -p$ and d = p$d = p$ is equal to 45: Sigma_6 = frac62 [2a + (6-1)d] = 45$$\Sigma_6 = \frac{6}{2} [2a + (6-1)d] = 45$$ 3 [2(-p) + 5p] = 45$$3 [2(-p) + 5p] = 45$$ 3 [3p] = 45 implies 9p = 45 implies p = 5$$3 [3p] = 45 \implies 9p = 45 \implies p = 5$$ ### Step 3: Calculate target absolute term difference The absolute difference between the 20^textth$20^{\text{th}}$ and 15^textth$15^{\text{th}}$ terms of any A.P. depends strictly on the common difference: |A_20 - A_15| = |(a + 19p) - (a + 14p)| = 5p$$|A_{20} - A_{15}| = |(a + 19p) - (a + 14p)| = 5p$$ 5p = 5(5) = 25$5p = 5(5) = 25$ ### Pattern Recognition The series sequence frac12 + frac16 + frac112 + dots$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \dots$ is the well-known telescoping series sum frac1n(n+1)$\sum \frac{1}{n(n+1)}$. Its sum to n$n$ terms is identically given by fracnn+1$\frac{n}{n+1}$ without requiring manual re-derivation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula Telescopic series decomposition via method of differences: frac1(k+2)(k+3) = frac1k+2 - frac1k+3$$\frac{1}{(k+2)(k+3)} = \frac{1}{k+2} - \frac{1}{k+3}$$ ### Core Logic Given: a_n = fracn^2+5n+64 = frac(n+2)(n+3)4$$a_n = \frac{n^2+5n+6}{4} = \frac{(n+2)(n+3)}{4}$$ Therefore, the reciprocal term is: frac1a_k = frac4(k+2)(k+3) = 4 left[ frac1k+2 - frac1k+3 right]$$\frac{1}{a_k} = \frac{4}{(k+2)(k+3)} = 4 \left[ \frac{1}{k+2} - \frac{1}{k+3} \right]$$ ### Step 1: Compute the Partial Sum S_n = sum_k=1^n frac1a_k = 4 sum_k=1^n left( frac1k+2 - frac1k+3 right)$$S_n = \sum_{k=1}^{n} \frac{1}{a_k} = 4 \sum_{k=1}^{n} \left( \frac{1}{k+2} - \frac{1}{k+3} \right)$$ Expanding the sum terms: S_n = 4 left[ left(frac13 - frac14right) + left(frac14 - frac15right) + dots + left(frac1n+2 - frac1n+3right) right]$$S_n = 4 \left[ \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \dots + \left(\frac{1}{n+2} - \frac{1}{n+3}\right) \right]$$ All intermediate terms cancel out: S_n = 4 left[ frac13 - frac1n+3 right] = 4 left[ fracn+3 - 33(n+3) right] = frac4n3(n+3)$$S_n = 4 \left[ \frac{1}{3} - \frac{1}{n+3} \right] = 4 \left[ \frac{n+3 - 3}{3(n+3)} \right] = \frac{4n}{3(n+3)}$$ ### Step 2: Calculate for n = 2025 For n = 2025$n = 2025$: S_2025 = frac4 times 20253 times (2025 + 3) = frac4 times 20253 times 2028$$S_{2025} = \frac{4 \times 2025}{3 \times (2025 + 3)} = \frac{4 \times 2025}{3 \times 2028}$$ We need to find 507 times S_2025$507 \times S_{2025}$: 507 times S_2025 = 507 times frac4 times 20253 times 2028$$507 \times S_{2025} = 507 \times \frac{4 \times 2025}{3 \times 2028}$$ Notice that 2028 = 4 times 507$2028 = 4 \times 507$: 507 times S_2025 = 507 times frac4 times 20253 times (4 times 507) = frac20253 = 675$$507 \times S_{2025} = 507 \times \frac{4 \times 2025}{3 \times (4 \times 507)} = \frac{2025}{3} = 675$$ ### Pattern Recognition Always look for arithmetic factor groupings at the end of large number sequence questions in JEE. Here recognizing 2028 = 4 times 507$2028 = 4 \times 507$ avoids large multi-digit multiplication. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula Sum of interior angles of an n$n$-sided polygon: S_n = (n - 2) times 180^circ$$S_n = (n - 2) \times 180^\circ$$ Sum of an Arithmetic Progression: S_n = fracn2 left[ 2a + (n-1)d right]$$S_n = \frac{n}{2} \left[ 2a + (n-1)d \right]$$ ### Core Logic The angles form an AP with common difference d = 6^circ$d = 6^\circ$. The largest angle is the last term: T_n = 219^circ$T_n = 219^\circ$. a + (n-1)6 = 219 implies a = 219 - 6n + 6 = 225 - 6n$$a + (n-1)6 = 219 \implies a = 219 - 6n + 6 = 225 - 6n$$ ### Step 1: Set up the sum equation Equating the two forms for the sum of angles: fracn2 left[ 2a + (n-1)6 right] = (n - 2) times 180$$\frac{n}{2} \left[ 2a + (n-1)6 \right] = (n - 2) \times 180$$ Substitute a = 225 - 6n$a = 225 - 6n$: fracn2 left[ 2(225 - 6n) + 6n - 6 right] = 180n - 360$$\frac{n}{2} \left[ 2(225 - 6n) + 6n - 6 \right] = 180n - 360$$ fracn2 left[ 450 - 12n + 6n - 6 right] = 180n - 360$$\frac{n}{2} \left[ 450 - 12n + 6n - 6 \right] = 180n - 360$$ fracn2 left[ 444 - 6n right] = 180n - 360$$\frac{n}{2} \left[ 444 - 6n \right] = 180n - 360$$ n(222 - 3n) = 180n - 360$$n(222 - 3n) = 180n - 360$$ 222n - 3n^2 = 180n - 360$$222n - 3n^2 = 180n - 360$$ 3n^2 - 42n - 360 = 0$$3n^2 - 42n - 360 = 0$$ ### Step 2: Solve the Quadratic Equation Divide by 3: n^2 - 14n - 120 = 0$$n^2 - 14n - 120 = 0$$ (n - 20)(n + 6) = 0$$(n - 20)(n + 6) = 0$$ Since number of sides n$n$ must be positive, n = 20$n = 20$. ### Pattern Recognition Always remember that any interior angle of a convex polygon must be less than 180^circ$180^\circ$. Let's check the smallest angle for n=20$n=20$: a = 225 - 120 = 105^circ$a = 225 - 120 = 105^\circ$, which is completely valid. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series