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The series 1 + 3 + 7 + 13 + 21 + … is not an AP and not a GP, so neither formula applies. But the gaps between its terms, 2, 4, 6, 8, form an AP, and that is enough to find any term and the sum. This is the method of differences. This post shows it for gaps in an AP, gaps in a GP, and fractions that telescope, with every answer checked against a direct sum.
Where does this fit in the syllabus?
The JEE (Advanced) 2026 brochure lists the "sum of the first n natural numbers, sums of squares and cubes of the first n natural numbers". The JEE (Main) 2026 syllabus lists arithmetic and geometric progressions. Neither names the method of differences; it is the tool that turns an unfamiliar series into those standard sums. Check the current syllabus for your exam year.
Our sequences and series guide uses telescoping to derive the sum of squares. Here we use those sums as tools.
How do you find the nth term when the gaps form an AP?
Example 1. Find the nth term and the sum of 1 + 3 + 7 + 13 + 21 + ….
Write the series twice, the second time shifted one place:
S = 1 + 3 + 7 + 13 + … + Tₙ
S = 1 + 3 + 7 + … + Tₙ₋₁ + Tₙ
Subtracting gives 0 = 1 + (2 + 4 + 6 + … to n − 1 terms) − Tₙ. The bracket is 2(1 + 2 + … + (n − 1)) = n(n − 1). So Tₙ = n² − n + 1. Check: T₅ = 25 − 5 + 1 = 21.
Now sum it with standard results:
Sₙ = Σk² − Σk + n = n(n + 1)(2n + 1)/6 − n(n + 1)/2 + n.
Over a common denominator of 6, the bracket is (n + 1)(2n + 1) − 3(n + 1) + 6 = 2n² + 4, so Sₙ = n(n² + 2)/3. Check: S₅ = 5 × 27/3 = 45, and 1 + 3 + 7 + 13 + 21 = 45.
What if the gaps form a GP?
Example 2. Find the nth term and the sum of 2 + 5 + 11 + 23 + 47 + ….
The gaps are 3, 6, 12, 24, a GP with ratio 2. So
Tₙ = 2 + 3(1 + 2 + 4 + … to n − 1 terms) = 2 + 3(2ⁿ⁻¹ − 1) = 3·2ⁿ⁻¹ − 1.
Check: T₅ = 48 − 1 = 47. Summing, Sₙ = 3(1 + 2 + … + 2ⁿ⁻¹) − n = 3(2ⁿ − 1) − n. Check: S₅ = 3 × 31 − 5 = 88, and 2 + 5 + 11 + 23 + 47 = 88.
How do fractions telescope?
If you can write each term as f(k) − f(k + 1), the middle terms cancel in pairs and only the ends survive.
Example 3. Find 1/(1·3) + 1/(3·5) + … + 1/((2n − 1)(2n + 1)).
Each term is (1/2)[1/(2k − 1) − 1/(2k + 1)]. Adding from k = 1 to n leaves (1/2)[1 − 1/(2n + 1)] = n/(2n + 1). Check with n = 2: 1/3 + 1/15 = 6/15 = 2/5, which matches n/(2n + 1) = 2/5. As n grows, the sum approaches 1/2.
Example 4. Find the sum of 1/(k(k + 1)(k + 2)) from k = 1 to n.
Each term is (1/2)[1/(k(k + 1)) − 1/((k + 1)(k + 2))]. The sum telescopes to (1/2)[1/2 − 1/((n + 1)(n + 2))] = 1/4 − 1/(2(n + 1)(n + 2)). Check with n = 1: 1/4 − 1/12 = 1/6, which is 1/(1·2·3).
Products telescope too. Since k(k + 1)(k + 2)(k + 3) − (k − 1)k(k + 1)(k + 2) = 4k(k + 1)(k + 2), the sum of k(k + 1)(k + 2) from 1 to n is n(n + 1)(n + 2)(n + 3)/4. For n = 2: 6 + 24 = 30, and 2 × 3 × 4 × 5/4 = 30.
Which mistakes cost the most marks?
Using n gaps instead of n − 1.
Assuming the gaps form an AP without checking. Work out a second row of gaps if the first is not obviously regular.
Dropping the factor 1/2 when splitting into partial fractions.
Losing a surviving end term after telescoping.
Not checking n = 1 and n = 2, which catches nearly all of these.
A practice routine
Explaining each step to yourself is what turns a worked example into a method you can reuse. In Chi, Bassok, Lewis, Reimann and Glaser (1989), the students who learned well from worked examples generated many self-explanations and related each step to the underlying principles; weaker students did not.
Write one sentence per step when you study a solved series: why the shift, why n − 1 terms, why that partial fraction.
Check every answer at n = 1, 2 and 5.
Sort before solving. For a mixed set of eight series, label each "gaps in AP", "gaps in GP", "telescoping" or "AP times GP". The last kind needs our arithmetico-geometric series method instead.
When you cannot see how to split a term, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, find the nth term of 3 + 8 + 15 + 24 + … and check it against the fifth term.
