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Mathematical Induction (HSC Maths Extension 1, 2026) quiz

10 questions. Pick an answer and you'll see why right away. Options are shuffled each attempt, and the review at the end links each missed question to the dot point to revise.

  1. Which four parts make up the standard structure of a proof by mathematical induction?

  2. In a proof that the statement P(n)P(n) holds, what does the inductive hypothesis assume?

  3. Proving ∑i=1ni2=n(n+1)(2n+1)6\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}, what is the base case value at n=1n = 1 for both sides?

  4. In proving ∑i=1n(2i−1)=n2\sum_{i=1}^{n} (2i-1) = n^2, the (k+1)(k+1)-th term added to the sum is:

  5. Proving 7n−17^n - 1 is divisible by 66, the hypothesis gives 7k=6M+17^k = 6M + 1. What is 7k+1−17^{k+1} - 1 in factored form?

  6. Proving 5n+35^n + 3 is divisible by 44, what is the base case n=1n = 1 value of 5n+35^n + 3?

  7. Proving 2n>n22^n > n^2 for n≥5n \ge 5 by induction, why must the base case be n=5n = 5 rather than n=1n = 1?

  8. Proving 9n−19^n - 1 is divisible by 88, the hypothesis is 9k=8M+19^k = 8M + 1. What is 9k+1−19^{k+1} - 1 in factored form?

  9. Proving ∑i=1n2i=2n+1−2\sum_{i=1}^{n} 2^i = 2^{n+1} - 2, adding the term 2k+12^{k+1} to (2k+1−2)(2^{k+1} - 2) gives:

  10. For the sequence a1=1a_1 = 1, an+1=3an+2a_{n+1} = 3 a_n + 2, applying the recursion to the hypothesis ak=2⋅3k−1−1a_k = 2 \cdot 3^{k-1} - 1 gives ak+1a_{k+1} equal to:

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