Use the Principle of Mathematical Induction t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(N/A) Let $P(n)$ be the statement $d_{n} = \frac{2}{n!}$ for $n \in N$.Step $1$: For $n=1$,$d_{1} = \frac{2}{1!} = \frac{2}{1} = 2$. This matches the

Vedclass · Accepted Answer

Vedclass · Answer

(N/A) Let $P(n)$ be the statement $d_{n} = \frac{2}{n!}$ for $n \in N$.
Step $1$: For $n=1$,$d_{1} = \frac{2}{1!} = \frac{2}{1} = 2$. This matches the given $d_{1}=2$. So,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$d_{k} = \frac{2}{k!}$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$d_{k+1} = \frac{2}{(k+1)!}$.
Given $d_{k+1} = \frac{d_{k}}{k+1}$.
Substituting the assumption $d_{k} = \frac{2}{k!}$,we get:
$d_{k+1} = \frac{2/k!}{k+1} = \frac{2}{k!(k+1)} = \frac{2}{(k+1)!}$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.

Use the Principle of Mathematical Induction to show that for a sequence $d_{1}, d_{2}, d_{3}, \ldots$ defined by $d_{1}=2$ and $d_{k}=\frac{d_{k-1}}{k}$ for all $k \geq 2$,the general term is $d_{n}=\frac{2}{n!}$ for all $n \in N$.

Similar Questions

Using mathematical induction,the numbers $a_n$ are defined by:
$a_0 = 1, a_{n+1} = 3n^2 + n + a_n, (n \geq 0)$.
Then,$a_n$ is equal to:

For what natural numbers $n \in N$,is the inequality $2^n > n+1$ valid?

Prove the following by using the principle of mathematical induction for all $n \in N:$
$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \ldots + \frac{1}{(2n-1)(2n+1)} = \frac{n}{2n+1}$

For all $n \ge 1,$ prove that $1^{2}+2^{2}+3^{2}+4^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}$

Prove the following by using the principle of mathematical induction for all $n \in N$:
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module