Q.Find the sum of the series (33−23)+(53−43)+(73−63)+… to
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What Does "Sum of a Series" Even Mean?
Imagine you're standing at point 0 and you take a step of 1 metre forward. Then a step of half a metre. Then a quarter metre. Then an eighth. And you keep going, each step half the size of the previous one.
After 1 step: you're at 1 metre.
After 2 steps: at 1.5 metres.
After 3 steps: at 1.75 metres.
After 4 steps: at 1.875 metres.
After 5 steps: at 1.9375 metres.
You notice something: you're getting closer and closer to 2 metres, but you never quite reach it. If you could take infinitely many steps, would you ever get to exactly 2 metres? This is the heart of what a series is — adding up infinitely many numbers and asking: does this sum settle down to a finite value?
A series is just the sum of the terms of a sequence. If the sequence is a1,a2,a3,…, then the series is a1+a2+a3+….
The Precise Definition
Let’s formalise this. Suppose we have an infinite sequence of numbers:
a1,a2,a3,a4,…
We want to make sense of the infinite sum:
a1+a2+a3+a4+…
We can't just "add infinitely many things" directly — that's not a finite operation. So mathematicians do something clever: they look at partial sums.
Define:
S1=a1
S2=a1+a2
S3=a1+a2+a3
⋮
Sn=a1+a2+⋯+an
Sn is called the nth partial sum — it's the sum of the first n terms.
Now, the series is said to converge (or have a sum) if the sequence of partial sums S1,S2,S3,… approaches some finite number S as n gets larger and larger. In that case, we write:
∑k=1∞ak=S
If the partial sums don't settle down to a finite number — they either grow without bound or oscillate forever — the series diverges and has no finite sum.
∑k=1∞ak=limn→∞SnwhereSn=∑k=1nak
The Walking Example, Formalised
Our sequence of steps: 1,21,41,81,…
The partial sums:
S1=1
S2=1+21=1.5
S3=1+21+41=1.75
S4=1+21+41+81=1.875
You can prove (and we will later) that:
Sn=2−2n−11
As n→∞, 2n−11→0, so Sn→2. Therefore:
∑k=1∞2k−11=2
The infinite sum equals exactly 2 — even though you never "reach" it after any finite number of steps, the limit of the process is 2.
Two Classic Examples to Build Intuition
1. The Harmonic Series (Diverges)
1+21+31+41+51+…
This one is tricky. The terms get smaller and smaller, but the partial sums grow without bound — just very slowly. S100≈5.18, S1000≈7.48, S1000000≈14.39. It never stops growing. This series diverges.
Just because terms get smaller does NOT mean the series converges. The harmonic series is the classic counterexample.
2. A Geometric Series (Converges) …
Concept: Sum of Series – Each term is a difference of cubes of consecutive odd and even numbers.
Step 1 – General term
The k-th pair is ((2k+1)3−(2k)3), for k=1,2,…,n.
Step 2 – Simplify the difference
Using a3−b3=(a−b)(a2+ab+b2) with a=2k+1, b=2k:
a−b=1, so
(2k+1)3−(2k)3=(2k+1)2+(2k+1)(2k)+(2k)2
=(4k2+4k+1)+(4k2+2k)+4k2
=12k2+6k+1.
Step 3 – Sum to n terms
Sn=∑k=1n(12k2+6k+1)=12⋅6n(n+1)(2n+1)+6⋅2n(n+1)+n
=2n(n+1)(2n+1)+3n(n+1)+n
=n[2(n+1)(2n+1)+3(n+1)+1] …
Sum to n terms is Sn=n(4n2+9n+6); sum to 10 terms is 4960.
The general term is the difference of consecutive odd/even cubes:
tn=(2n+1)3−(2n)3.
Expanding with (a+1)3−a3=3a2+3a+1 where a=2n:
tn=3(2n)2+3(2n)+1=12n2+6n+1.
(i) Sum to n terms
Sn=∑k=1n(12k2+6k+1)=12⋅6n(n+1)(2n+1)+6⋅2n(n+1)+n. …
- KEAM 2026Set eng-2026-04184 marksMCQQ.If ∑k=1nlog10(5k)=66log10(5), then the value of n is equal to (A) 8 (B) 9 (C) 10 (D) 11 (E) 12
›Reveal solutionSolution
2n(n+1)=66⇒n(n+1)=132⇒n=11.
log10(5k)=klog105, so
∑k=1nklog105=2n(n+1)log105.
Setting this equal to 66log105: …
- KEAM 2025Set eng-2025-04284 marksMCQQ.The sum of the geometric series 3+12+48+ ...up to 10 terms is (A) 10233 (B) 10243 (C) 5113 (D) 5123 (E) 2153
›Reveal solutionSolution
Factor 3: terms are 3(1,2,4,…), a GP with r=2. Sum of 10 terms =3⋅2−1210−1=10233.
Simplify: 12=23, 48=43, so the series is
3(1+2+4+⋯),
a geometric series with first term 3 and ratio 2. …
- KEAM 2024Set eng-2024-06054 marksMCQQ.The sum upto n terms of 1+61+6+111+… is (A) 51[5n+1] (B) 51[5n+1+1] (C) 51[5n+1−1] (D) 61[6n+1] (E) 71[7n+1−1]
›Reveal solutionSolution
Rationalise each term to get a telescoping difference; the sum collapses to 51(5n+1−1).
The k-th term is 5k−4+5k+11 (denominators 1,6,11,…).
Rationalise: 5k−4+5k+11=(5k+1)−(5k−4)5k+1−5k−4=55k+1−5k−4.
Summing k=1 to n telescopes: …
- KEAM 2024Set eng-2024-06074 marksMCQQ.The value of the sum k=0∑48(k+1)(k+2)1 is equal to (A) 5051 (B) 4951 (C) 5049 (D) 4948 (E) 4950
›Reveal solutionSolution
Split into partial fractions and telescope. …
- KEAM 2022Set eng-2022-P2-B14 marksMCQQ.If nth term of a series is n+(−1)n−1, n=1,2,3,…, then the sum of first 40 terms of the series is (A) 810 (B) 820 (C) 821 (D) 819 (E) 780
›Reveal solutionSolution
The sum of the first 40 terms is 820.
Concept and Intuition
Split the term into its arithmetic part n and its alternating part (−1)n−1 and sum each separately.
Step-by-Step Solution
- an=n+(−1)n−1.
- ∑n=140n=240⋅41=820.
- ∑n=140(−1)n−1=1−1+1−⋯=0 (20 cancelling pairs).
- Total =820+0=820.
Common Mistakes …
- KEAM 2022Set eng-2022-P2-B14 marksMCQQ.Let tn=n1∑k=1n(nk)2 for n=1,2,3,…. Then t10 is equal to (A) 6007 (B) 100231 (C) 600209 (D) 20011 (E) 20077
›Reveal solutionSolution
t10=20077.
Concept and Intuition
Factor out constants and use ∑k=1nk2=6n(n+1)(2n+1).
Step-by-Step Solution
- tn=n1∑k=1nn2k2=n31∑k=1nk2.
- =n31⋅6n(n+1)(2n+1)=6n2(n+1)(2n+1).
- For n=10: 6⋅10011⋅21=600231.
- Simplify: 600231=20077. …
- KEAM 2021Set eng-2021-P2-B14 marksMCQQ.The value of k=5∑36k2−k1 is (A) 367 (B) 91 (C) 92 (D) 121 (E) 365
›Reveal solutionSolution
The sum equals 92.
Concept and Intuition
The term k(k−1)1 splits by partial fractions into k−11−k1, producing a telescoping series where all interior terms cancel, leaving only the first and last.
Step-by-Step Solution
- k2−k1=k(k−1)1=k−11−k1.
- Sum from k=5 to 36 telescopes to 5−11−361=41−361. …
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