Let a_1 = 1 and for n geq 1 , a_n+1 = frac12 a_n + fracn^2 - 2n - 1n^2 (n + 1)^2 . Then left|sum_n=1^inftyleft(a_n - frac2n^2right)right| is equal

Numerical Answer Type:
Enter a numerical value Answer: 2 to 2 +4 marks

Solution & Explanation

### Related Formula Partial fraction decomposition for telescopic summing: frac2n^2 - (n+1)^2 + 1 dotsdots text structures directly cancel in series expansions. ### Core Logic Given recurrence: a_n+1 - frac12a_n = fracn^2 - 2n - 1n^2(n+1)^2 Rewrite the numerator to split the fraction: n^2 - 2n - 1 = 2n^2 - (n^2 + 2n + 1) = 2n^2 - (n+1)^2 a_n+1 - frac12a_n = frac2n^2 - (n+1)^2n^2(n+1)^2 = frac2(n+1)^2 - frac1n^2 ### Step 1: Telescope generation Multiply both sides by appropriate powers of 2 to create a cancelling chain: For n=1: a_2 - frac12a_1 = frac22^2 - frac11^2 For n=2: multiply by 2 Rightarrow 2left[a_3 - frac12a_2 = frac23^2 - frac12^2right] Rightarrow 2a_3 - a_2 = frac2 times 23^2 - frac22^2 Wait, let's look at a cleaner telescopic scaling: a_n+1 - frac2(n+1)^2 = frac12 left(a_n - frac2n^2right). ### Step 2: Identify Geometric Progression Let V_n = a_n - frac2n^2. The recurrence gives V_n+1 = frac12 V_n. This proves V_n is a geometric progression with common ratio r = 1/2. First term V_1 = a_1 - frac21^2 = 1 - 2 = -1. ### Step 3: Infinite Summation We need left| sum_n=1^infty left( a_n - frac2n^2 right) right| = left| sum_n=1^infty V_n right|. Since V_n is an infinite GP: S_infty = fracV_11 - r = frac-11 - 1/2 = frac-11/2 = -2 Taking absolute value: |-2| = 2 ### Pattern Recognition When dealing with rational fraction recurrences A_n+1 - k A_n = f(n) - k f(n-1), immediately substitute 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

Reference Study Guides

Sequences And Series

More Sequences and Series Previous-Year Questions — Page 5

Q59 jee_main_2025_24_jan_evening Arithmetic Progression Sum
In an arithmetic progression, if S_40=1030 and S_12=57, then S_30-S_10 is equal to: [cite: 3294, 3295, 3296]
  • A. 510
  • B. 515
  • C. 525
  • D. 505

Solution

### Related Formula Sum of first n terms of an Arithmetic Progression: S_n = fracn2[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_12 = frac122[2a + 11d] = 57 Rightarrow 2a + 11d = 9.5 ### Step 1: Solve for a and d Subtract the second equation from the first : (2a + 39d) - (2a + 11d) = 51.5 - 9.5 28d = 42 Rightarrow d = frac4228 = frac32 = 1.5 Substitute d = 1.5 back to find a: 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 Write out the formula for the target subtraction : S_30 - S_10 = frac302[2a + 29d] - frac102[2a + 9d] = 15(2a + 29d) - 5(2a + 9d) = 30a + 435d - 10a - 45d = 20a + 390d [cite: 3964, 3965] Substitute the values of a and d : = 20(-3.5) + 390(1.5) = -70 + 585 = 515 ### Pattern Recognition Notice that S_30 - S_10 represents the \sum of terms from T_11 to T_30, which can also be formulated as 20 times A_20.5, saving algebraic steps if calculated symmetrically. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q60 jee_main_2025_24_jan_evening Arithmetico-Geometric Progression
If 7=5+frac17(5+alpha)+frac17^2(5+2alpha)+frac17^3(5+3alpha)+dotsdotsinfty, then the value of alpha is: [cite: 3301, 3302]
  • A. 1
  • B. frac67
  • C. 6
  • D. frac17

Solution

### Related Formula Sum of an infinite geometric progression: S_infty = fraca1-r quad textfor |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 ### Step 1: Shift and Subtract Multiply the equation by the common ratio frac17 and shift it by one position : frac17S = frac57 + frac5+alpha7^2 + frac5+2alpha7^3 + dots infty Subtract this from the original equation: S - frac17S = 5 + left(frac5+alpha-57right) + left(frac5+2alpha-(5+alpha)7^2right) + dots frac67S = 5 + fracalpha7 + fracalpha7^2 + fracalpha7^3 + dots ### Step 2: Sum the Infinite Geometric Series Apply the infinite GP formula to the terms involving alpha : frac67S = 5 + fracalpha7left(frac11 - frac17right) = 5 + fracalpha7left(frac76right) = 5 + fracalpha6 Given that S = 7 : frac67(7) = 5 + fracalpha6 Rightarrow 6 = 5 + fracalpha6 1 = fracalpha6 Rightarrow alpha = 6 ### Pattern Recognition Standard trick for infinite AGPs: Multiply by the common ratio 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
Q54 jee_main_2025_24_jan_morning Sum to n terms of Special Series
Let S_n = frac12 + frac16 + frac112 + frac120 + dots up to n terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is sqrt2026S_2025, then the absolute difference between 20^textth and 15^textth terms of the A.P. is :
  • A. 25
  • B. 90
  • C. 20
  • D. 45

Solution

### Related Formula The general term for the provided series is: T_k = frac1k(k+1) = frac1k - frac1k+1 This sets up a standard telescoping summation sequence. ### Core Logic Express the sum 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 = 1 - frac12026 = frac20252026 ### Step 1: Compute the boundary expression value Substitute S_2025 into the expression value: sqrt2026 cdot S_2025 = sqrt2026 cdot frac20252026 = sqrt2025 = 45 ### Step 2: Apply Arithmetic Progression Summation The sum of the first 6 terms of the A.P. with a = -p and d = p is equal to 45: Sigma_6 = frac62 [2a + (6-1)d] = 45 3 [2(-p) + 5p] = 45 3 [3p] = 45 implies 9p = 45 implies p = 5 ### Step 3: Calculate target absolute term difference The absolute difference between the 20^textth and 15^textth terms of any A.P. depends strictly on the common difference: |A_20 - A_15| = |(a + 19p) - (a + 14p)| = 5p 5p = 5(5) = 25 ### Pattern Recognition The series sequence frac12 + frac16 + frac112 + dots is the well-known telescoping series sum frac1n(n+1). Its sum to n terms is identically given by fracnn+1 without requiring manual re-derivation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q63 jee_main_2025_28_jan_evening Telescopic Series Summation
For positive integers n, if 4a_n=(n^2+5n+6) and S_n=sum_k=1^nleft(frac1a_kright) then the value of 507 S_2025 is:
  • A. 540
  • B. 1350
  • C. 675
  • D. 135

Solution

### Related Formula Telescopic series decomposition via method of differences: frac1(k+2)(k+3) = frac1k+2 - frac1k+3 ### Core Logic Given: a_n = fracn^2+5n+64 = 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] ### 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) Expanding the sum terms: S_n = 4 left[ left(frac13 - frac14right) + left(frac14 - frac15right) + dots + left(frac1n+2 - frac1n+3right) 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) ### Step 2: Calculate for n = 2025 For n = 2025: S_2025 = frac4 times 20253 times (2025 + 3) = frac4 times 20253 times 2028 We need to find 507 times S_2025: 507 times S_2025 = 507 times frac4 times 20253 times 2028 Notice that 2028 = 4 times 507: 507 times S_2025 = 507 times frac4 times 20253 times (4 times 507) = frac20253 = 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 avoids large multi-digit multiplication. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q73 jee_main_2025_28_jan_evening Arithmetic Progression Applications
The interior angles of a polygon with n sides, are in an A.P. with common difference 6^circ If the largest interior angle of the polygon is 219^circ, then n is equal to
Numerical Answer. Answer: 20 to 20

Solution

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

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