Summation — Natural numbers, Σr², Σr³
Sequences & Series
14
JEE Qs
8%
Hard
60
min
Memorize the formulas for Σr, Σr², Σr³ and practice adapting them when the summation does not start from 1 or when the general term needs simplification.
🧮 Key Formulas
✅ Key Points for JEE
- 1The formulas for Σr, Σr², Σr³ are valid only when the summation starts from r=1 and consists of consecutive natural numbers up to 'n'.
- 2If a summation starts from a number 'm' (where m > 1), express it as a difference: Σ(r=m to n) T_r = Σ(r=1 to n) T_r - Σ(r=1 to m-1) T_r.
- 3For a general series, first identify the r-th term (T_r) of the series. If T_r is a polynomial in 'r', expand it and then apply the individual summation formulas term by term using the linearity of summation (Σ(A_r + B_r) = ΣA_r + ΣB_r).
- 4Always simplify the resulting algebraic expression in 'n' to its most factorized form; this often helps in matching options or further calculations.
⚠️ Common Mistakes
- ✕Applying the formulas directly when the series does not start from r=1 or is not consecutive, without adjusting the limits.
- ✕Making algebraic errors while expanding the r-th term or simplifying the final expression involving 'n'.
- ✕Incorrectly remembering or misapplying the specific formulas, especially confusing Σr² and Σr³.
- ✕Forgetting to expand a product in the r-th term (e.g., r(r+1)) before applying summation formulas.
📝 Practice Questions
See allQ12.The remainder, when 7103 is divided by 23 , is equal to : (1) 6 (2) 17 (3) 9 (4) 14
Q22.Let a1, a2, … , a2024 be an Arithmetic Progression such that a1 + (a5 + a10 + a15 + … + a2020) + a2024 = 2233. Then a1 + a2 + a3 + … + a2024 is equal to _______ 1 2 3 , then α is equal to ________ (3x + t = 5eα ( 85 )
Q1. Let a1, a2, a3, … be a G.P. of increasing positive terms. If a1a5 = 28 and a2 + a4 = 29, then a6 is equal to: (1) 628 (2) 812 (3) 526 (4) 784 = 0. If x(1) = 1, then x ( 12 ) is :
Q15.If ∑nr=1 Tr = (2n−1)(2n+1)(2n+3)(2n+5)64 , then limn→∞∑nr=1 ( Tr1 ) (1) 0 (2) 23 (3) 1 (4) 13
Q13.Suppose that the number of terms in an A.P. is 2k, k ∈N . If the sum of all odd terms of the A.P. is 40 , the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to : (1) 6 (2) 5 (3) 8 (4) 4 y+2
Q1. If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to (1) −1080 (2) −1020 (3) −1200 (4) −120
NCERT Chapters
- Class 11 Maths Ch 9: Sequences and Series