Special Series and Summation

• Standard summation formulas: Σk = n(n+1)/2, Σk² = n(n+1)(2n+1)/6, Σk³ = [n(n+1)/2]².
• Telescoping series — writing terms as differences and cancelling.
• Method of differences for summing non-standard sequences.

Level 1 — Three standard formulas

Sum of first n natural numbers: Σk = 1 + 2 + 3 + … + n = n(n+1)/2.

Sum of squares: Σk² = 1² + 2² + … + n² = n(n+1)(2n+1)/6.

Sum of cubes: Σk³ = 1³ + 2³ + … + n³ = [n(n+1)/2]² = (Σk)².

Remarkable identity: (Σk³) = (Σk)² — the sum of cubes equals the square of the sum of natural numbers.

Derived sums:

• Sum of first n even numbers: 2 + 4 + 6 + … + 2n = n(n+1).
• Sum of first n odd numbers: 1 + 3 + 5 + … + (2n−1) = n².
• Σk(k+1) = Σk² + Σk = n(n+1)(2n+1)/6 + n(n+1)/2 = n(n+1)(n+2)/3.

Level 2 — Telescoping series and method of differences

Telescoping series: Write general term as f(k) − f(k−1) (or f(k+1) − f(k)). When summed, intermediate terms cancel and only boundary terms remain.

Example template: If Tₖ = 1/(k(k+1)) = 1/k − 1/(k+1), then Σ Tₖ (k=1 to n) = 1 − 1/(n+1) = n/(n+1).

Method of differences: For a sequence whose differences form a known pattern (AP, GP, etc.):

1. Let Sₙ = T₁ + T₂ + … + Tₙ.
2. Shift: Sₙ = T₁ + T₂ + … + Tₙ₋₁ + Tₙ (written one step shifted).
3. Subtract to get Tₙ in terms of differences.
4. Sum the resulting simpler series.

Vₙ method (partial fractions for products): Tₙ = 1/[n(n+1)(n+2)] can be telescoped using identity: 1/[n(n+1)(n+2)] = (1/2)[1/(n(n+1)) − 1/((n+1)(n+2))].

Polynomial sums: Sₙ = Σaₖ where Tₙ is a polynomial in n of degree d. Express Tₙ in terms of combinations C(n,1), C(n,2), … and sum using ΣC(k,r) = C(n+1,r+1).

JEE pattern — Vₙ method generalised: For Tₙ involving product of consecutive integers, write Tₙ = (1/m)·[f(n+1) − f(n)] where f(n) is product of (m−1) consecutive integers starting at n.

NCERT spotlight

Verify: 1³ + 2³ + 3³ + 4³ = 100 = (1+2+3+4)² = 10² = 100. ✓ The telescoping trick is essential for JEE — identify if Tₙ = Aₙ − Aₙ₋₁ for some sequence Aₙ. Then Σ Tₙ = Aₙ − A₀. Method of differences: if differences of differences are constant, it is a polynomial sequence.

Find Σ 1/(k(k+2)) for k = 1 to n.

Step 1 — Partial fractions: 1/(k(k+2)) = (1/2)[1/k − 1/(k+2)].
Step 2 — Write out terms:
         k=1: (1/2)[1/1 − 1/3]
         k=2: (1/2)[1/2 − 1/4]
         k=3: (1/2)[1/3 − 1/5]
         ...
         k=n: (1/2)[1/n − 1/(n+2)]
Step 3 — Sum (telescoping — most terms cancel):
         Sₙ = (1/2)[(1 + 1/2) − (1/(n+1) + 1/(n+2))].
Step 4 — Sₙ = (1/2)[3/2 − (1/(n+1) + 1/(n+2))].
Step 5 — Sₙ = 3/4 − (1/2)[1/(n+1) + 1/(n+2)].

Find Sₙ = 1·2 + 2·3 + 3·4 + … + n(n+1).

Step 1 — General term: Tₖ = k(k+1) = k² + k.
Step 2 — Sum: Sₙ = Σk² + Σk = n(n+1)(2n+1)/6 + n(n+1)/2.
Step 3 — Factor n(n+1): Sₙ = n(n+1)[(2n+1)/6 + 1/2] = n(n+1)[(2n+1+3)/6].
Step 4 — Sₙ = n(n+1)(2n+4)/6 = n(n+1)·2(n+2)/6 = n(n+1)(n+2)/3.
Step 5 — Check n=3: 1·2 + 2·3 + 3·4 = 2 + 6 + 12 = 20. Formula: 3·4·5/3 = 20. ✓

1. Find 1² + 2² + 3² + … + 10².
2. Evaluate 1³ + 2³ + 3³ + … + 6³.
3. Find Σ 1/(k(k+1)) for k = 1 to 100.
4. Sum the series: 1·2·3 + 2·3·4 + 3·4·5 + … to n terms.
5. Stretch: Prove that Σk³ = (Σk)² using the formula for Σk³ and Σk, and verify the identity holds for n = 4.

Open the Practice tab for graded questions on Sequences & Series — Special Series.

• Use the platform-native live simulation or PhET-style tool for this topic (number line, Venn, physics playground, molecule builder, sensor dashboard, etc.).
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