Further Mathematics · OCR H245 · Topic 4.06

Series

Summation of series, method of differences.

  • 2 sub-topics
  • 1 in Year 12, 1 in Year 13
  • 28 practice questions
  • 2 animated lessons
Practice questions (28) Lessons Worksheets coming soon All OCR Further Mathematics topics

What the specification says

Every sub-topic OCR can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 4.06aSummation of seriesY12
  2. 4.06bMethod of differencesY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 28 questions in this topic, free, no account needed.

Further algebra and functions

Easy

Standard sums · Year 12 · 1 mark

  1. What is ∑r=112r\displaystyle\sum_{r=1}^{12} r?
    (1)

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  1. ∑r=1nr\displaystyle\sum_{r=1}^{n} r=12n(n+1){}= \tfrac{1}{2}n(n + 1).CC: 78

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Further algebra and functions

Medium

Standard sums · Year 12 · 3 marks

  1. Find ∑r=119(4r2+5)\sum_{r=1}^{19} (4r^2 + 5).
    (3)

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  1. Split and use standard results.4×16n(n+1)(2n+1)4 \times \tfrac{1}{6}n(n + 1)(2n + 1)+5n{}+ 5n
  2. n=19n = 19
  3. 99759975

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Further algebra and functions

Hard

Sum in terms of n · Year 12 · 3 marks

  1. Show that ∑r=1nr(r+3)\sum_{r=1}^{n} r(r + 3)=16n(n+1)(2n+10){}= \tfrac{1}{6}n(n + 1)(2n + 10).
    (3)
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  1. Split the sum into standard results.∑r2+3∑r\sum r^2 + 3\sum r
  2. 16n(n+1)(2n+1)\tfrac{1}{6}n(n + 1)(2n + 1)+32n(n+1){}+ \tfrac{3}{2}n(n + 1)
  3. Factorise.16n(n+1)(2n+1+9)\tfrac{1}{6}n(n + 1)(2n + 1 + 9)

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