Summation of Series — Telescoping + Σr Σr² Σr³
Sequences & Series
38
JEE Qs
30%
Hard
90
min
For summation problems, always start by identifying the general r-th term (T_r) and then decide if it simplifies to a standard sum, requires a telescoping decomposition, or a combination of methods.
🧮 Key Formulas
✅ Key Points for JEE
- 1The core idea of a telescoping series is to express the r-th term T_r as a difference of two consecutive terms of a sequence, i.e., T_r = V_r - V_{r+k} or V_{r+k} - V_r, such that intermediate terms cancel out upon summation.
- 2Master the standard summation formulas for Σr, Σr², Σr³ as they are frequently used directly or after simplification of the general term.
- 3For telescoping sums, the key step is to use partial fraction decomposition, algebraic manipulation (e.g., multiplying and dividing by a suitable term), or rationalization to transform the general term T_r into the desired difference form.
- 4Carefully write out the first few and last few terms of the series after expressing T_r in difference form to correctly observe the cancellation pattern and identify the terms that remain.
- 5Be vigilant about the limits of summation and potential index shifts. Ensure the general term T_r is correctly identified for the r-th term of the series, not just a pattern.
⚠️ Common Mistakes
- ✕Incorrectly applying standard sum formulas by using 'n' when the actual upper limit of summation is different, or misremembering the formulas themselves.
- ✕Errors in partial fraction decomposition or algebraic manipulation when trying to convert T_r into the V_r - V_{r+k} form.
- ✕Failing to identify the correct cancellation pattern in telescoping sums, leading to incorrect remaining terms or signs, especially with multi-term differences.
- ✕Not finding the general r-th term (T_r) correctly for more complex series, which makes applying any summation method impossible.
📝 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 Mathematics Ch 9: Sequences and Series