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CodeVID-M11-15-CT
Principle of Mathematical Induction — Full Chapter Test
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- This is a full-length test covering the whole chapter — every topic is included.
- 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.
Induction needs a base case and the:
- A.converse
- B.inductive step
- C.contrapositive
- D.negation
2.
$1+2+\cdots+n$ equals:
- A.$n^2$
- B.$\dfrac{n(n+1)}{2}$
- C.$2^n-1$
- D.$n(n+1)$
3.
$3^{2n}-1$ is divisible by:
- A.$3$
- B.$6$
- C.$8$
- D.$9$
4.
Bernoulli's inequality is:
- A.$(1+x)^n \le 1+nx$
- B.$(1+x)^n \ge 1+nx$
- C.$(1+x)^n = 1+nx$
- D.$(1+x)^n \ge nx$
5.
$2^n > n^2$ first holds at:
- A.$n=2$
- B.$n=3$
- C.$n=4$
- D.$n=5$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
6.
State the two steps of mathematical induction.
7.
Verify the base case of $1+2+\cdots+n=\dfrac{n(n+1)}{2}$.
8.
Write $f(k+1)$ for $f(n)=7^n-3^n$.
9.
Give the base case for $2^n > n^2$.
Section C — Short Answer (3 marks)
4 × 3 = 12 marks
10.
Prove the inductive step for $1+3+\cdots+(2n-1)=n^2$.
11.
Show $3^{2(k+1)}-1=8(9m+1)$ given $3^{2k}-1=8m$.
12.
Show that $2^k > k$ implies $2^{k+1} > k+1$.
13.
Explain why a missing base case can invalidate an induction.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
14.
Prove by induction that $1^2+2^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}$.
15.
Prove by induction that $(1+x)^n \ge 1+nx$ for all $n\in\mathbb{N}$, $x>-1$.
Answer Key
Section A — Multiple Choice Questions
- (B) inductive step
- (B) $\dfrac{n(n+1)}{2}$
- (C) $8$
- (B) $(1+x)^n \ge 1+nx$
- (D) $n=5$
Section B — Short Answer (2 marks)
- Base case $P(1)$, and $P(k)\Rightarrow P(k+1)$ for all $k\ge1$.
- $n=1$: $1=\dfrac{1\cdot2}{2}=1$; true.
- $f(k+1)=7\cdot7^k-3\cdot3^k$.
- $P(5)$: $32>25$.
Section C — Short Answer (3 marks)
- $k^2+(2k+1)=(k+1)^2$.
- $9(8m+1)-1=72m+8=8(9m+1)$.
- $2^{k+1}=2\cdot2^k>2k\ge k+1$ for $k\ge1$.
- Without a true base case the chain never starts, so nothing is proved even if the step holds.
Section D — Long Answer (5 marks)
- Base $n=1$ true; adding $(k+1)^2$ to the $k$-case gives $\dfrac{(k+1)(k+2)(2k+3)}{6}$; by PMI true for all $n$.
- Base $n=1$: equality; multiplying by $1+x>0$ gives $(1+x)^{k+1}\ge1+(k+1)x+kx^2\ge1+(k+1)x$; by PMI true for all $n$.
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