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

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Sequences And Series

More Sequences and Series Previous-Year Questions — Page 8

Q25 jee_main_2024_30_january_evening Arithmetic Progression
Let S_n be the sum to n-terms of an arithmetic progression 3, 7, 11, dots If 40 lt left(frac6n(n + 1)sum_k=1^nS_kright) lt 42 , then n equals
Numerical Answer. Answer: 9 to 9

Solution

### Related Formula textSum of AP: S_k = frack2 [2a + (k - 1)d] sum_k=1^n k^2 = fracn(n+1)(2n+1)6 sum_k=1^n k = fracn(n+1)2 ### Core Logic For the arithmetic progression 3, 7, 11, dots First term a = 3, Common difference d = 4. The sum of the first k terms is: S_k = frack2 (2(3) + (k - 1)4) = frack2 (6 + 4k - 4) = frack2 (4k + 2) = 2k^2 + k ### Step 1: Finding the Sum of Sums Now compute the sum sum_k=1^n S_k: sum_k=1^n S_k = sum_k=1^n (2k^2 + k) = 2sum_k=1^n k^2 + sum_k=1^n k = 2 left( fracn(n+1)(2n+1)6 right) + fracn(n+1)2 = n(n+1) left[ frac2(2n+1)6 + frac12 right] = n(n+1) left[ frac2n+13 + frac12 right] = n(n+1) left[ frac4n + 2 + 36 right] = fracn(n+1)(4n + 5)6 ### Step 2: Resolving the Inequality Substitute this sum into the given expression: frac6n(n+1) sum_k=1^n S_k = frac6n(n+1) cdot fracn(n+1)(4n+5)6 = 4n + 5 We are given the bounds: 40 lt 4n + 5 lt 42 35 lt 4n lt 37 8.75 lt n lt 9.25 Since n must be an integer (representing the number of terms), the only valid integer is n = 9. ### Pattern Recognition Evaluating a 'sum of sums' for an AP effectively requires applying the Sigma k^2 and Sigma k standard formulas to the generic S_n quadratic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series
Q2 jee_main_2024_30_jan_morning Sum of n terms of AP
Let S_n denote the sum of first n terms an arithmetic progression. If S_20 = 790 and S_10 = 145, then S_15 - S_5 is:
  • A. 395
  • B. 390
  • C. 405
  • D. 410

Solution

### Related Formula S_n = fracn2[2a + (n-1)d] ### Core Logic Using the sum formula for an AP: S_20 = frac202[2a + 19d] = 790 10[2a + 19d] = 790 2a + 19d = 79 quad dots (1) S_10 = frac102[2a + 9d] = 145 5[2a + 9d] = 145 2a + 9d = 29 quad dots (2) ### Step 1: Solving for a and d Subtracting (2) from (1): 10d = 50 Rightarrow d = 5 Substituting d=5 into (2): 2a + 9(5) = 29 Rightarrow 2a = 29 - 45 = -16 a = -8 ### Step 2: Evaluating the required expression We need to find S_15 - S_5: S_15 - S_5 = frac152[2a + 14d] - frac52[2a + 4d] Substituting 2a = -16 and d = 5: = frac152[-16 + 70] - frac52[-16 + 20] = frac152[54] - frac52[4] = 15 times 27 - 5 times 2 = 405 - 10 = 395 ### Pattern Recognition When two sums of an AP are given, immediately set up the linear equations in terms of a and d. Solve for them, and substitute directly into the target expression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series
Q30 jee_main_2024_30_jan_morning Special Series
Let alpha = 1^2 + 4^2 + 8^2 + 13^2 + 19^2 + 26^2 + dots upto 10 terms and beta = sum_n=1^10 n^4. If 4alpha - beta = 55k + 40, then k is equal to
Numerical Answer. Answer: 353 to 353

Solution

### Related Formula textMethod of differences for a sequence: V_n - V_n-1 = T_n ### Core Logic The base terms inside the squares form a sequence: 1, 4, 8, 13, 19, 26 dots The differences between consecutive terms are: 3, 4, 5, 6, 7 dots Since the first differences are an Arithmetic Progression, the general term of the inner sequence is a quadratic in n: T_n = an^2 + bn + c. Using n=1: 1 = a + b + c Using n=2: 4 = 4a + 2b + c Using n=3: 8 = 9a + 3b + c Solving this system: (4a + 2b + c) - (a + b + c) = 3 Rightarrow 3a + b = 3 (9a + 3b + c) - (4a + 2b + c) = 4 Rightarrow 5a + b = 4 Subtracting these gives: 2a = 1 Rightarrow a = 1/2. Then 3(1/2) + b = 3 Rightarrow b = 3/2. Finally 1/2 + 3/2 + c = 1 Rightarrow c = -1. Inner sequence T_n = frac12n^2 + frac32n - 1. ### Step 1: Calculating alpha structure The series is alpha = sum_n=1^10 (T_n)^2. 4alpha = sum_n=1^10 4left(fracn^2 + 3n - 22right)^2 = sum_n=1^10 (n^2 + 3n - 2)^2 Expand the squared trinomial: (n^2 + 3n - 2)^2 = n^4 + 9n^2 + 4 + 6n^3 - 4n^2 - 12n = n^4 + 6n^3 + 5n^2 - 12n + 4 ### Step 2: Applying given target relation We are given beta = sum_n=1^10 n^4. So, 4alpha - beta = sum_n=1^10 (n^4 + 6n^3 + 5n^2 - 12n + 4) - sum_n=1^10 n^4 4alpha - beta = sum_n=1^10 (6n^3 + 5n^2 - 12n + 4) ### Step 3: Calculating summation limits Evaluate each standard summation up to n=10: sum n^3 = (10 times 11 / 2)^2 = 55^2 = 3025 sum n^2 = (10 times 11 times 21) / 6 = 385 sum n = (10 times 11) / 2 = 55 sum 4 = 40 4alpha - beta = 6(3025) + 5(385) - 12(55) + 40 = 18150 + 1925 - 660 + 40 = 19455 We are given 4alpha - beta = 55k + 40. 19455 = 55k + 40 19415 = 55k k = frac1941555 = 353 ### Pattern Recognition Recognizing arithmetic progressions in the first-order differences immediately specifies a quadratic general term An^2+Bn+C. Expanding and cancelling highest-order summation terms drastically simplifies standard sums. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series
Q9 jee_main_2024_31_jan_evening Arithmetic and Geometric Progression
Let 2^mathrmnd, 8^mathrmth and 44^mathrmth, terms of a non-constant A.P. be respectively the 1^mathrmst, 2^mathrmnd and 3^mathrmrd terms of G.P. If the first term of A.P. is 1 then the sum of first 20 terms is equal to
  • A. 980
  • B. 960
  • C. 990
  • D. 970

Solution

### Related Formula S_n = fracn2[2a + (n-1)d] textIf p, q, r text are in G.P. then q^2 = pr ### Core Logic Let the A.P. be a, a+d, a+2d, dots Given a=1, the 2^textnd, 8^textth, and 44^textth terms are: T_2 = 1 + d T_8 = 1 + 7d T_44 = 1 + 43d These terms are in G.P., so: (1+7d)^2 = (1+d)(1+43d) 1 + 14d + 49d^2 = 1 + 44d + 43d^2 6d^2 - 30d = 0 implies 6d(d - 5) = 0 Since it is a non-constant A.P., d neq 0, so d = 5. Sum of first 20 terms: S_20 = frac202[2(1) + (20-1)5] S_20 = 10[2 + 19(5)] = 10[2 + 95] = 970 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series
Q13 jee_main_2024_31_jan_morning Method of Differences
The sum of the series frac11 - 3 cdot 1^2 + 1^4 + frac21 - 3 cdot 2^2 + 2^4 + frac31 - 3 cdot 3^2 + 3^4 + dots up to 10 terms is
  • A. frac45109
  • B. -frac45109
  • C. frac55109
  • D. -frac55109

Solution

### Core Logic General term T_r = fracrr^4 - 3r^2 + 1. Factorize the denominator: r^4 - 3r^2 + 1 = (r^4 - 2r^2 + 1) - r^2 = (r^2 - 1)^2 - r^2 = (r^2 - r - 1)(r^2 + r - 1) ### Step 1: Partial Fractions T_r = fracr(r^2 - r - 1)(r^2 + r - 1) Notice that (r^2 + r - 1) - (r^2 - r - 1) = 2r. T_r = frac12 left[ frac2r(r^2 - r - 1)(r^2 + r - 1) right] = frac12 left[ frac1r^2 - r - 1 - frac1r^2 + r - 1 right] ### Step 2: Telescoping Sum Sum S = sum_r=1^10 T_r. The terms will telescope because the second term for r is identical to the first term for r+1. (Let v_r = r^2 - r - 1, then v_r+1 = (r+1)^2 - (r+1) - 1 = r^2 + 2r + 1 - r - 1 - 1 = r^2 + r - 1). S = frac12 left[ frac11^2 - 1 - 1 - frac110^2 + 10 - 1 right] S = frac12 left[ frac1-1 - frac1109 right] = frac12 left[ -1 - frac1109 right] = -frac55109 ### Pattern Recognition Expressions like r^4 + kr^2 + 1 can be factorized by completing the square to create a difference of two squares. This setup invariably leads to a telescoping series. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

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