Algebra – Mathematical Induction Quiz - ICSE Class 11 Mathematics
Free practice quiz: Algebra – Mathematical Induction (ICSE Class 11 Mathematics). 57 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
Play all 57 questions freeSample questions
Let $P(n)$ be the statement: $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + n(n+1) = \frac{n(n+1)(n+2)}{3}$ for all positive integers $n$. If $P(k)$ is true, what is $P(k+1)$?
- $\frac{k(k+1)(k+2)}{3} + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}$
- $\frac{k(k+1)(k+2)}{3} + k(k+1) = \frac{k(k+1)(k+2)}{3}$
- $\frac{k(k+1)(k+2)}{3} + (k+1)(k+2) = \frac{k(k+1)(k+2)}{3}$
- $\frac{k(k+1)(k+2)}{3} + (k+1)(k+2) = \frac{k(k+2)(k+3)}{3}$
Show answer
Answer: A. $\frac{k(k+1)(k+2)}{3} + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}$
To prove $P(k+1)$, we assume $P(k)$ is true and show that $P(k+1)$ is also true. $P(k)$ states that the sum up to $k$ terms is $\frac{k(k+1)(k+2)}{3}$. The sum for $P(k+1)$ includes the $(k+1)$-th term, which is $(k+1)((k+1)+1) = (k+1)(k+2)$. Therefore, $P(k+1)$ is the sum of $P(k)$ and the $(k+1)$-th term: $\frac{k(k+1)(k+2)}{3} + (k+1)(k+2)$. We then need to show this equals the formula for $n=k+1$, which is $\frac{(k+1)((k+1)+1)((k+1)+2)}{3} = \frac{(k+1)(k+2)(k+3)}{3}$. The correct option represents this step.
Let $P(n)$ be the statement: $1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + n(n+1)(n+2) = \frac{n(n+1)(n+2)(n+3)}{4}$. If $P(k)$ is true, what is the expression for $P(k+1)$?
- $\frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3) = \frac{(k+1)(k+2)(k+3)(k+4)}{4}$
- $\frac{k(k+1)(k+2)(k+3)}{4} + k(k+1)(k+2) = \frac{(k+1)(k+2)(k+3)(k+4)}{4}$
- $\frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3) = \frac{k(k+2)(k+3)(k+4)}{4}$
- $\frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3) = \frac{k(k+1)(k+2)(k+3)}{4}$
Show answer
Answer: A. $\frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3) = \frac{(k+1)(k+2)(k+3)(k+4)}{4}$
Assuming $P(k)$ is true, the sum of the first $k$ terms is $\frac{k(k+1)(k+2)(k+3)}{4}$. The $(k+1)$-th term in the series is $(k+1)((k+1)+1)((k+1)+2) = (k+1)(k+2)(k+3)$. To get $P(k+1)$, we add this $(k+1)$-th term to the sum of the first $k$ terms: $\frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3)$. We then need to show this equals the formula for $n=k+1$, which is $\frac{(k+1)((k+1)+1)((k+1)+2)((k+1)+3)}{4} = \frac{(k+1)(k+2)(k+3)(k+4)}{4}$. Option 1 correctly represents this inductive step.
Consider the statement: $1^2 - 2^2 + 3^2 - 4^2 + \dots + (-1)^{n-1}n^2 = (-1)^{n-1} \frac{n(n+1)}{2}$. What is the base case for this statement?
- n=1
- n=2
- n=0
- n=-1
Show answer
Answer: A. n=1
The statement is for all positive integers $n$. The smallest positive integer is $n=1$. For $n=1$, the left side is $(-1)^{1-1}1^2 = (-1)^0 \cdot 1 = 1$. The right side is $(-1)^{1-1} \frac{1(1+1)}{2} = (-1)^0 \frac{1 \cdot 2}{2} = 1 \cdot 1 = 1$. Since $1=1$, the statement is true for $n=1$.
Let $P(n)$ be the statement: $\sum_{i=1}^n \frac{1}{i(i+1)} = \frac{n}{n+1}$. What is the $(k+1)$-th term in the summation?
- $\frac{1}{k(k+1)}$
- $\frac{1}{(k+1)(k+2)}$
- $\frac{k}{k+1}$
- $\frac{k+1}{k+2}$
Show answer
Answer: B. $\frac{1}{(k+1)(k+2)}$
The summation is given by $\sum_{i=1}^n \frac{1}{i(i+1)}$. The $(k+1)$-th term corresponds to the value of the expression when $i = k+1$. Substituting $i = k+1$ into the general term $\frac{1}{i(i+1)}$, we get $\frac{1}{(k+1)((k+1)+1)} = \frac{1}{(k+1)(k+2)}$.
Which of the following is a typical domain for mathematical induction proofs?
- All real numbers.
- All positive integers.
- All rational numbers.
- All complex numbers.
Show answer
Answer: B. All positive integers.
Mathematical induction is primarily used to prove statements that hold for all positive integers (or sometimes for all non-negative integers, or for all integers greater than or equal to some specific integer). Its recursive nature relies on the well-ordered property of these sets.