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 .
Part a
Prove that = 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, =1=. Assume =. Then =+(2k+1)=+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
= 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 .
- Identify exactly which summand comes next.
Compare your own reasoning. No automatic examination marks are awarded.