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How do you use the standard summation formulae for sums of powers to evaluate more complicated series?

The standard results for the sum of r, r squared and r cubed, using them to sum polynomial expressions in r, splitting sums by linearity, and adjusting limits.

A focused answer to the OCR A-Level Further Mathematics A content on the summation of series, covering the standard formulae for the sum of r, r squared and r cubed, using linearity to split a sum of a polynomial in r, evaluating the resulting expression, and adjusting the limits when a sum does not start at one.

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  1. What this dot point is asking
  2. The standard results
  3. Splitting a sum by linearity
  4. Factorising the result
  5. Adjusting the limits
  6. Try this

What this dot point is asking

OCR wants you to know and use the standard summation results for βˆ‘r\sum r, βˆ‘r2\sum r^2 and βˆ‘r3\sum r^3, to sum a polynomial expression in rr by splitting it with the linearity of summation, to factorise and simplify the resulting expression, and to adjust the limits when a sum starts above 11 by writing it as a difference of two sums from 11.

The standard results

These three formulae, together with βˆ‘1=n\sum 1 = n, are the toolkit for the whole topic. They are quoted in the OCR formulae booklet, but you should know them well enough to apply them instantly.

Notice the neat fact that βˆ‘r3=(βˆ‘r)2\sum r^3 = \left(\sum r\right)^2, which OCR sometimes asks you to verify or use.

Splitting a sum by linearity

Summation distributes over addition and pulls out constants, so a sum of a polynomial in rr breaks into separate standard sums. This is the routine first move on every summation question.

Factorising the result

After applying the formulae you almost always have a common factor of n(n+1)n(n+1) (and a fractional coefficient). Taking out the largest common factor and tidying the bracket gives the clean factorised answer that the mark scheme expects. The reliable routine is: first write each standard sum in full, then identify the common factor shared by every term (it is usually 16n(n+1)\tfrac{1}{6}n(n+1) when a βˆ‘r2\sum r^2 is present, or 14n2(n+1)2\tfrac{1}{4}n^2(n+1)^2 when the sum is dominated by βˆ‘r3\sum r^3), pull it out front, and simplify the bracket that remains into a single polynomial. Expanding the bracket, collecting like terms, and then refactoring often reveals a tidy linear or quadratic factor such as (2n+5)(2n + 5) or (n+5)(n + 5). Always finish by checking your factorised formula against a small value of nn (for example n=1n = 1 or n=2n = 2) computed directly from the original sum; if the two agree, the algebra is almost certainly correct, and this quick check catches most slips.

Adjusting the limits

When a sum does not start at r=1r = 1, rewrite it as the difference of two sums that do. This lets you use the standard formulae, which are stated from 11.

The standard sums underpin the method of differences and are themselves provable by induction, tying the whole series and proof strand together.

Try this

Q1. Find βˆ‘r=1n(r+2)\displaystyle\sum_{r=1}^{n} (r + 2). [2 marks]

  • Cue. βˆ‘r+βˆ‘2=12n(n+1)+2n=12n(n+5)\sum r + \sum 2 = \tfrac{1}{2}n(n+1) + 2n = \tfrac{1}{2}n(n+5).

Q2. State the value of βˆ‘r=110r3\displaystyle\sum_{r=1}^{10} r^3. [2 marks]

  • Cue. 14(10)2(11)2=14β‹…100β‹…121=3025\tfrac{1}{4}(10)^2(11)^2 = \tfrac{1}{4}\cdot 100\cdot 121 = 3025.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20185 marksFind βˆ‘r=1n(r2+3r)\displaystyle\sum_{r=1}^{n} (r^2 + 3r) in terms of nn, giving your answer as a single factorised expression.
Show worked answer β†’

Split by linearity (M1): βˆ‘r=1n(r2+3r)=βˆ‘r=1nr2+3βˆ‘r=1nr\displaystyle\sum_{r=1}^{n}(r^2 + 3r) = \sum_{r=1}^{n} r^2 + 3\sum_{r=1}^{n} r.

Use the standard results (M1, A1): =16n(n+1)(2n+1)+3β‹…12n(n+1)= \tfrac{1}{6}n(n+1)(2n+1) + 3\cdot\tfrac{1}{2}n(n+1).

Factor out 16n(n+1)\tfrac{1}{6}n(n+1) (M1): =16n(n+1)[(2n+1)+9]=16n(n+1)(2n+10)= \tfrac{1}{6}n(n+1)\left[(2n+1) + 9\right] = \tfrac{1}{6}n(n+1)(2n+10) (A1).

Simplify: =13n(n+1)(n+5)= \tfrac{1}{3}n(n+1)(n+5).

Markers reward splitting the sum, quoting the standard formulae, taking out the common factor, and simplifying.

OCR 20226 marksFind βˆ‘r=n+12nr3\displaystyle\sum_{r=n+1}^{2n} r^3 in terms of nn, using the standard result for βˆ‘r=1nr3\displaystyle\sum_{r=1}^{n} r^3.
Show worked answer β†’

Write the sum from n+1n+1 to 2n2n as a difference of two sums from 11 (M1): βˆ‘r=n+12nr3=βˆ‘r=12nr3βˆ’βˆ‘r=1nr3\displaystyle\sum_{r=n+1}^{2n} r^3 = \sum_{r=1}^{2n} r^3 - \sum_{r=1}^{n} r^3.

Apply βˆ‘r=1Nr3=14N2(N+1)2\displaystyle\sum_{r=1}^{N} r^3 = \tfrac{1}{4}N^2(N+1)^2 (M1): =14(2n)2(2n+1)2βˆ’14n2(n+1)2= \tfrac{1}{4}(2n)^2(2n+1)^2 - \tfrac{1}{4}n^2(n+1)^2 (A1).

Simplify the first term (A1): 14β‹…4n2(2n+1)2=n2(2n+1)2\tfrac{1}{4}\cdot 4n^2(2n+1)^2 = n^2(2n+1)^2.

So the sum is n2(2n+1)2βˆ’14n2(n+1)2n^2(2n+1)^2 - \tfrac{1}{4}n^2(n+1)^2 (M1), which can be left in this form or factored as 14n2[4(2n+1)2βˆ’(n+1)2]\tfrac{1}{4}n^2\left[4(2n+1)^2 - (n+1)^2\right] (A1).

Markers reward writing the partial sum as a difference, applying the cube formula to both, simplifying the upper term, and a correct final expression.

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