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Exam-style practice

A sum, an induction proof and an application

Choose your method, connect the parts and explain your conclusions. These original questions include written reasoning and sketches as well as exact answer checks.

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Before you start

Proof by induction. Read the relevant method, then decide which parts you can solve without hints.

Check a misconception first · Try a connected problem

Read an original example question

Let Sn=1+3+⋯+(2n−1)S_n=1+3+\cdots+(2n-1).

Part a

Prove that SnS_{n} = n2n^{2} for every integer n ≥ 1.

Part b

Hence find the number of terms needed for a sum of 100.

Part c

A proposed induction step adds 2k−1. Explain the mistake.

Model solution and review criteria

Part a

At n=1, S1S_{1}=1=121^{2}. Assume SkS_{k}=k2k^{2}. Then Sk+1S_{k+1}=SkS_{k}+(2k+1)=k2k^{2}+2k+1=(k+1)². The base case and arbitrary-step implication prove the claim.

  • Check the base case.
  • State the arbitrary k hypothesis.
  • Use the next summand and complete the square.
  • Conclude over the stated domain.

Part b

n2n^{2} = 100 and n is a positive integer, so n = 10.

  • Use the proven formula and discard a negative term count.

Part c

The hypothesis already includes 2k−1. The next term is 2(k+1)−1=2k+1. Adding the previous term does not produce Sk+1S_{k+1}.

  • Identify exactly which summand comes next.

Compare your own reasoning. No automatic examination marks are awarded.