Question types

Mathematical Induction question types

68 questions across 5 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.

68
Questions
5
Question groups
5
Question types
Sample Questions

Mathematical Induction questions

One sample from each question group in this chapter. Select any group above to see the full set with answer keys.

If $10^\text{n} + 3 \times 4^{\text{n}+2}+\lambda$ is divisible by 9 for all $\text{n}\in\text{N},$ then the least positive integer value of $\lambda$ is 
  1. 5
  2. 3
  3. 7
  4. 1
View full solution
A student was asked to prove a statement p(n) by induction. He proved p(K + 1) is true whenever p(k) is true for all $\text{k}>5\in\text{N}$ and also p(5) is true. On the basis of this he could conclude that p(n) is true.
  1. For all $\text{n}\in\text{N}$
  2. For all n > 5
  3. For all $\text{n}\geq5$
  4. For all n > 5
View full solution
Prove that $\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+...+\cos(\alpha+(\text{n}-1)\beta)\\=\frac{\cos\Big\{\alpha+\big(\frac{\text{n}-1}{2}\big)\beta\Big\}\sin\big(\frac{\text{n}\beta}{2}\big)}{\sin\frac{\beta}{2}}$ For all $\text{n}\in\text{N}.$
View full solution
A sequence x1, x2, x3, ... is defined by letting x1 = 2 and $\text{x}_{\text{k}}=\frac{\text{x}_{\text{k}}-1}{\text{n}}$ for all natural numbers k, $\text{k}\geq2.$ Show that $\text{x}_{\text{n}}=\frac{2}{\text{n}!}$ for all $\text{n}\in\text{N}.$
View full solution
Prove the following by the principle of mathematical induction:
$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(\text{3n-1)(3n+2)}}=\frac{\text{n}}{\text{6n}+4}$
View full solution
Prove the following by the principle of mathematical induction:
$\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{(\text{4n-1)(4n+3)}}=\frac{\text{n}}{3(\text{4n}+3)}$
View full solution

Generate a Mathematical Induction paper free

Pick question groups from the list above, set marks and difficulty, and export a branded PDF with step-by-step answer keys. First 3 chapters free — no signup.

Download App