Series and the Method of Differences
Standard summation formulas
Further Maths uses standard results for sums of series (which can be proved by induction):
Σ(r=1 to n) r = ½n(n + 1)
Σ(r=1 to n) r² = (1/6)n(n + 1)(2n + 1)
Σ(r=1 to n) r³ = ¼n²(n + 1)² = [Σr]²
These let you sum polynomial series without adding term by term.
Using the formulas
- Σ is linear:
Σ(ar² + br + c) = aΣr² + bΣr + cn(Σ of a constant c over n terms = cn). - To sum from a value other than 1, use
Σ(r=1 to n) − Σ(r=1 to m−1).
Example: Σ(r=1 to n)(2r + 3) = 2·½n(n+1) + 3n = n(n+1) + 3n = n² + 4n.
The method of differences
Some series can be summed if each term can be written as the difference of two consecutive terms of another sequence:
if uᵣ = f(r) − f(r+1) then Σuᵣ = f(1) − f(n+1)
Most middle terms cancel (a "telescoping" sum), leaving only the first and last. Partial fractions are often used to write a fraction as such a difference.
Example: 1/[r(r+1)] = 1/r − 1/(r+1), so the sum telescopes to 1 − 1/(n+1).
Maclaurin series (if in your spec)
Functions can be written as a power series:
f(x) = f(0) + f′(0)x + [f″(0)/2!]x² + [f‴(0)/3!]x³ + …
giving series for eˣ, sin x, cos x, ln(1+x), used for approximations.
Worked example
Find Σ(r=1 to n) r² for n = 4.
- Formula: (1/6)(4)(5)(9) = (1/6)(180) = 30. Check: 1 + 4 + 9 + 16 = 30 ✓.
Common mistakes
- Forgetting Σc (a constant) over n terms is cn, not c.
- Getting the standard formulas slightly wrong — learn them precisely.
- In the method of differences, mis-cancelling the telescoping terms.
Exam tips
- Split sums using linearity, then apply the standard formulas.
- For the method of differences, use partial fractions to expose the telescoping structure.
- Substitute a small n to check your final expression.
Key facts to remember
- Σr = ½n(n+1), Σr² = (1/6)n(n+1)(2n+1), Σr³ = ¼n²(n+1)²; Σ is linear (Σc = cn).
- Method of differences: if uᵣ = f(r) − f(r+1), the sum telescopes to f(1) − f(n+1) (use partial fractions).
- Maclaurin series expands functions as power series for approximations.