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3.4 Sequences and Series (A-level only)

3.4 Sequences and Series (A-level only)

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Question 131
a.

Using nCr=n!r!(n−r)!{}^nC_r = \frac{n!}{r!(n-r)!}nCr​=r!(n−r)!n!​, show that nC4=n(n−1)(n−2)(n−3)24{}^nC_4 = \frac{n(n-1)(n-2)(n-3)}{24}nC4​=24n(n−1)(n−2)(n−3)​.

[2]
bi.

A researcher is selecting distinct plant species from a population of nnn available species for a DNA sequencing study. Show that the equation

2×nC4=15×nC2 2 \times {}^nC_4 = 15 \times {}^nC_2 2×nC4​=15×nC2​

simplifies to

n2−5n−84=0 n^2 - 5n - 84 = 0 n2−5n−84=0
[3]
bii.

Hence, solve the equation

2×nC4=15×nC2 2 \times {}^nC_4 = 15 \times {}^nC_2 2×nC4​=15×nC2​
[2]
Markscheme

3.4 Sequences and Series (A-level only) Questions

  1. A Level
  2. /Maths
  3. /3.4 Sequences and Series (A-level only)

231 exam-style questions on WJEC A Level Maths 3.4 Sequences and Series (A-level only), covering 3.4.1 Sequences and Series (A-level only), 3.4.2 Sequences and Series (A-level only), 3.4.3 Sequences and Series (A-level only), 3.4.4 Sequences and Series (A-level only), 3.4.5 Sequences and Series (A-level only), 3.4.6 Sequences and Series (A-level only), and 3.4.7 Sequences and Series (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.

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