Telescoping Series
Sequences & Series
14
JEE Qs
8%
Hard
60
min
Master the art of transforming the general term T_k into a difference f(k) - f(k+c) using partial fractions, rationalization, or other algebraic manipulation, as this is the key to all telescoping sums.
🧮 Key Formulas
✅ Key Points for JEE
- 1The core idea is to express the general term T_k as a difference of two consecutive (or c-apart) terms, i.e., T_k = f(k) - f(k+c) or T_k = f(k+c) - f(k).
- 2The most common technique to achieve the difference form for rational functions is partial fraction decomposition.
- 3For terms involving square roots in the denominator (e.g., 1/(sqrt(k)+sqrt(k+1))), rationalization is often used to transform them into a difference form.
- 4Carefully write out the first few and last few terms of the sum to identify the exact terms that cancel and which ones remain.
- 5Pay close attention to the index 'k' and the constant 'c' in 'f(k) - f(k+c)' to correctly determine the non-cancelling terms at the beginning and end of the series.
⚠️ Common Mistakes
- ✕Incorrectly decomposing the general term into the difference form, especially sign errors or constant factors.
- ✕Errors in identifying the remaining terms after cancellation, particularly when c > 1 (e.g., T_k = f(k) - f(k+2) will leave f(1), f(2) at the start and -f(n+1), -f(n+2) at the end).
- ✕Algebraic mistakes during partial fraction decomposition or rationalization.
- ✕Failing to adapt the decomposition technique to slightly different general terms (e.g., 1/(k(k+2)) vs 1/(k(k+1))).
📝 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