Vidaara.orgClass 11 · Mathematics
CodeVID-M11-15-PMI-01
The Principle of Mathematical Induction — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
A proof by induction requires the base case and the:
- A.converse
- B.inductive step
- C.contrapositive
- D.negation
2.
For a claim valid for all $n \ge 3$, the base case is:
- A.$P(1)$
- B.$P(2)$
- C.$P(3)$
- D.$P(0)$
3.
The inductive hypothesis assumes the statement holds for:
- A.$n = k$
- B.$n = k+1$
- C.all $n$
- D.$n = 1$ only
4.
Which alone proves $P(n)$ for all $n$?
- A.base case only
- B.inductive step only
- C.both together
- D.checking $5$ cases
5.
The $(k+1)$th odd number $2(k+1)-1$ equals:
- A.$2k-1$
- B.$2k$
- C.$2k+1$
- D.$2k+2$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
6.
State the two steps of the principle of mathematical induction.
7.
Verify the base case of $P(n): 1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}$.
8.
Write the inductive hypothesis for $P(n): 2^n > n$.
9.
Give the base case of a statement valid for all $n \ge 5$.
Section C — Short Answer (3 marks)
4 × 3 = 12 marks
10.
For $P(n): 1+3+\cdots+(2n-1)=n^2$, write $P(k+1)$ explicitly.
11.
Explain why a true base case but no inductive step proves only one statement.
12.
Give an example showing finitely many true cases need not prove a statement.
13.
In $P(n): 1+2+\cdots+n=\dfrac{n(n+1)}{2}$, simplify $\dfrac{k(k+1)}{2}+(k+1)$.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
14.
Prove by induction that $1+2+3+\cdots+n=\dfrac{n(n+1)}{2}$ for all $n\in\mathbb{N}$.
15.
Prove by induction that $1+3+5+\cdots+(2n-1)=n^2$ for all $n\in\mathbb{N}$.
Answer Key
Section A — Multiple Choice Questions
- (B) inductive step
- (C) $P(3)$
- (A) $n = k$
- (C) both together
- (C) $2k+1$
Section B — Short Answer (2 marks)
- Base case $P(1)$ true, and $P(k) \Rightarrow P(k+1)$ for all $k \ge 1$.
- $n=1$: LHS $=1$, RHS $=\dfrac{1\cdot2}{2}=1$; true.
- Assume $2^k > k$ for some fixed $k \ge 1$.
- $P(5)$.
Section C — Short Answer (3 marks)
- $1+3+\cdots+(2k-1)+(2k+1)=(k+1)^2$.
- It verifies just $P(1)$; nothing links it to later $n$.
- $n^2-n+41$ is prime for $n=1,\dots,40$ but composite at $n=41$.
- $\dfrac{(k+1)(k+2)}{2}$.
Section D — Long Answer (5 marks)
- Base $n=1$ true; assuming the $k$-case, adding $(k+1)$ gives $\dfrac{(k+1)(k+2)}{2}$; by PMI it holds for all $n$.
- Base $n=1$: $1=1^2$; assuming $k^2$, adding $(2k+1)$ gives $k^2+2k+1=(k+1)^2$; by PMI true for all $n$.
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