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 3

Q59 jee_main_2025_28_jan_morning Recurrence Relations and Summation
Let < a_n > be a sequence such that a_0 = 0, a_1 = frac12 and 2a_n + 2 = 5a_n + 1 - 3a_n, n = 0, 1, 2, 3, dots. Then sum_k = 1^100 a_k is equal to: (1) 3a_99 - 100 (2) 3a_100 - 100 (3) 3a_100 + 100 (4) 3a_99 + 100
  • A. 3a_99 - 100
  • B. 3a_100 - 100
  • C. 3a_100 + 100
  • D. 3a_99 + 100

Solution

### Related Formula Characteristic equation method for standard second-order linear homogeneous recurrence updates: 2x^2 - 5x + 3 = 0 ### Core Logic Solving the characteristic equation gives roots x = 1 and x = frac32. The general solution takes the form: a_n = A(1)^n + Bleft(frac32right)^n ### Step 1: Evaluating Sequence Parameters Using boundary conditions: For n = 0 implies A + B = 0 For n = 1 implies A + frac32B = frac12 Solving this simple linear system gives B = 1 and A = -1. Thus, the explicit sequence formula is: a_n = -1 + left(frac32right)^n ### Step 2: Summing the Target Range sum_k = 1^100 a_k = sum_k = 1^100 (-1) + sum_k = 1^100 left(frac32right)^k = -100 + fracfrac32left[left(frac32right)^100 - 1right]frac32 - 1 = -100 + 3left[left(frac32right)^100 - 1right] = 3a_100 - 100 ### Pattern Recognition Characteristic roots directly decouple second-order linear loop progressions into basic combinations of clean geometric progressions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series
Q61 jee_main_2025_28_jan_morning Arithmetic Progression Properties
Let T_r be the r^textth term of an A.P. If for some m, T_m = frac125, T_25 = frac120 and 20sum_r = 1^25 T_r = 13 then 5msum_r = m^2m T_r is equal to: (1) 112 (2) 126 (3) 98 (4) 142
  • A. 112
  • B. 126
  • C. 98
  • D. 142

Solution

### Related Formula Standard Arithmetic Progression summation template: S_n = fracn2[2a + (n-1)d] ### Core Logic Given structural constraints: T_25 = a + 24d = frac120 20 cdot frac252left[a + frac120right] = 13 implies a = frac1500 ### Step 1: Finding Parameters and Indices Substituting a = frac1500 back into a + 24d = frac120 gives d = frac1500. Using the formula for T_m: T_m = a + (m-1)d = frac1500 + fracm-1500 = frac125 implies m = 20 ### Step 2: Computing the Target Segment Sum For m = 20, the target expression becomes: 5(20) sum_r=20^40 T_r = 100 cdot frac212 [T_20 + T_40] Evaluating the values gives exactly 126. ### Pattern Recognition When a = d, the expressions simplify directly to basic multiples of the index position (T_n = n cdot d), cutting down calculation time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series
Q57 jee_main_2025_03_april_morning Method of Differences
The sum 1 + 3 + 11 + 25 + 45 + 71 + dots up to 20 terms, is equal to[cite: 580, 584]:
  • A. 7240
  • B. 7130
  • C. 6982
  • D. 8124

Solution

### Related Formula For series whose consecutive differences form an Arithmetic Progression (A.P.), the general term is given by a quadratic form: T_n = an^2 + bn + c ### Core Logic Analyze successive first-order differences of the terms [cite: 1293, 1294]: textSeries: 1, quad 3, quad 11, quad 25, quad 45, quad 71 [cite: 1293] textDifferences: 2, quad 8, quad 14, quad 20, quad 26 [cite: 1294] Since the differences grow uniformly by 6, they reside in an A.P. [cite: 1294] Thus, set up the general term system [cite: 1296, 1298]: - T_1 = a + b + c = 1 - T_2 = 4a + 2b + c = 3 - T_3 = 9a + 3b + c = 11 Solving the linear equations simultaneously yields [cite: 1299]: a = 3, quad b = -7, quad c = 5 [cite: 1299] ### Step 1: Summing the Series The general term is [cite: 1300]: T_n = 3n^2 - 7n + 5 [cite: 1300] Evaluate the summation for n=20 terms [cite: 1302]: S_20 = sum_n=1^20 (3n^2 - 7n + 5) = 3sum_n=1^20 n^2 - 7sum_n=1^20 n + sum_n=1^20 5 [cite: 1302] Substitute standard sequence formulas [cite: 1302]: S_20 = 3 cdot left(frac20 cdot 21 cdot 416right) - 7 cdot left(frac20 cdot 212right) + 5(20) [cite: 1302] = 8610 - 1470 + 100 = 7240 [cite: 1302] ### Pattern Recognition When first-order differences form a regular arithmetic line, the original function is exactly quadratic. Identify coefficients using small terms quickly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q66 jee_main_2025_03_april_morning Geometric Progression
Let a_1, a_2, a_3, ldots be a G.P. of increasing positive numbers[cite: 655]. If a_3a_5 = 729 and a_2 + a_4 = frac1114 [cite: 656], then 24(a_1 + a_2 + a_3) is equal to[cite: 657]:
  • A. 131
  • B. 130
  • C. 129
  • D. 128

Solution

### Related Formula For a geometric sequence configuration with first term a and common ratio r: a_n = a cdot r^n-1 ### Core Logic Convert information markers using parameter notations [cite: 1372, 1373, 1376]: a_3 a_5 = (ar^2)(ar^4) = a^2 r^6 = 729 implies ar^3 = 27 [cite: 1373, 1374] From second expression block data [cite: 1376]: a_2 + a_4 = ar + ar^3 = frac1114 [cite: 1376] Substitute ar^3 = 27 directly into the linear equation block [cite: 1376]: ar + 27 = frac1114 implies ar = frac1114 - 27 = frac34 [cite: 1376] ### Step 1: Finding parameters a and r Divide the calculated components to evaluate the ratio [cite: 1387]: fracar^3ar = frac273/4 implies r^2 = 36 implies r = 6 [cite: 1387] (Choose +6 because terms must stay strictly positive [cite: 655]). Find value for first base variable a [cite: 1389]: a(6) = frac34 implies a = frac18 [cite: 1389] ### Step 2: Sum configuration resolving Now compute targeted expansion expression value [cite: 1390]: 24(a_1 + a_2 + a_3) = 24(a + ar + ar^2) = 24a(1 + r + r^2) [cite: 1390] = 24 cdot left(frac18right) cdot (1 + 6 + 36) = 3 cdot 43 = 129 [cite: 1390, 1391] ### Pattern Recognition Product entries like a_3 a_5 = a_4^2 help identify the central term index value quickly in symmetric geometric progressions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q62 jee_main_2025_04_april_evening Telescoping Series
If the sum of the first 20 terms of the series frac 4 . 14 + 3 . 1 ^ 2 + 1 ^ 4 + frac 4 . 24 + 3 . 2 ^ 2 + 2 ^ 4 + frac 4 . 34 + 3 . 3 ^ 2 + 3 ^ 4 + frac 4 . 44 + 3 . 4 ^ 2 + 4 ^ 4 + dots is fracmathrmmmathrmn, where m and n are coprime, then mathrmm + mathrmn is equal to:
  • A. 423
  • B. 420
  • C. 421
  • D. 422

Solution

### Core Logic The general term T_r of the series can be written as: T_r = frac4rr^4 + 3r^2 + 4 Let's factorize the denominator by completing the square metric: r^4 + 3r^2 + 4 = (r^4 + 4r^2 + 4) - r^2 = (r^2 + 2)^2 - r^2 Using the difference of squares identity A^2 - B^2 = (A-B)(A+B): r^4 + 3r^2 + 4 = (r^2 - r + 2)(r^2 + r + 2) ### Step 1: Partial Fraction Decomposition Express T_r using partial fractions split: T_r = frac4r(r^2 - r + 2)(r^2 + r + 2) = 2 left[ frac1r^2 - r + 2 - frac1r^2 + r + 2 right] Notice that if we define V(r) = r^2 - r + 2, then V(r+1) = (r+1)^2 - (r+1) + 2 = r^2 + 2r + 1 - r - 1 + 2 = r^2 + r + 2. Thus, T_r = 2big[V(r) - V(r+1)big], which sets up a clear telescoping sum formulation. ### Step 2: Evaluating the Sum of 20 Terms Summing from r = 1 to 20: S_20 = sum_r=1^20 T_r = 2 sum_r=1^20 left[ frac1r^2 - r + 2 - frac1r^2 + r + 2 right] = 2 left[ left(frac12 - frac14right) + left(frac14 - frac18right) + dots + left(frac120^2 - 20 + 2 - frac120^2 + 20 + 2right) right] All sequential middle terms cancel completely, leaving only first and final values: S_20 = 2 left[ frac12 - frac1422 right] = 1 - frac1211 = frac210211 Since 210 and 211 are coprime, m = 210 and n = 211. ### Step 3: Calculating m + n Combining both values: m + n = 210 + 211 = 421 ### Pattern Recognition The polynomial factorization r^4 + a^2r^2 + b^4 is a frequent pattern in series problems. Always complete the square to break it into a product of quadratic expressions, which naturally yields a telescoping sequence. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

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