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FP1.7 - Series

Syllabus
2019
Topic
—
Level
AS

Sum finite series with the standard results

A finite sum can be evaluated by rewriting its summand as a polynomial in the index, splitting it term by term, and applying the standard results. Here nn is a positive integer.

Sum from r=1r=1 to nn Standard result
∑1\displaystyle\sum 1 nn
∑r\displaystyle\sum r n(n+1)2\displaystyle\frac{n(n+1)}2
∑r2\displaystyle\sum r^2 n(n+1)(2n+1)6\displaystyle\frac{n(n+1)(2n+1)}6
∑r3\displaystyle\sum r^3 (n(n+1)2)2\displaystyle\left(\frac{n(n+1)}2\right)^2

Summation is linear: constants may be taken outside and separate polynomial terms may be summed separately. Thus ∑r=1n(ar3+br2+cr+d)=a∑r=1nr3+b∑r=1nr2+c∑r=1nr+dn.\sum_{r=1}^{n}(ar^3+br^2+cr+d)=a\sum_{r=1}^{n}r^3+b\sum_{r=1}^{n}r^2+c\sum_{r=1}^{n}r+dn. Expand or factor the summand before choosing the formulas.

For the official model form, ∑r=1nr(r2+2)=∑r=1n(r3+2r)=(n(n+1)2)2+n(n+1).\sum_{r=1}^{n}r(r^2+2)=\sum_{r=1}^{n}(r^3+2r)=\left(\frac{n(n+1)}2\right)^2+n(n+1). Factoring gives n(n+1)4(n(n+1)+4).\frac{n(n+1)}4\bigl(n(n+1)+4\bigr). This preserves the equality for every positive integer nn, rather than checking only particular values.

For a different lower limit, define F(m)=∑r=1mf(r)F(m)=\sum_{r=1}^{m}f(r). Then ∑r=abf(r)=F(b)−F(a−1).\sum_{r=a}^{b}f(r)=F(b)-F(a-1). For example, with f(r)=r2+2f(r)=r^2+2, ∑r=410(r2+2)=F(10)−F(3)=(385+20)−(14+6)=385.\sum_{r=4}^{10}(r^2+2)=F(10)-F(3)=(385+20)-(14+6)=385.

Subtract through a−1a-1, not through aa. If a sum starts at r=0r=0, include the r=0r=0 term separately; in particular, ∑r=0nc=c(n+1)\sum_{r=0}^{n}c=c(n+1). The method of differences is not required in FP1, so do not replace this standard-results method with telescoping or partial-fraction differences.