Use the Principle of Mathematical Induction t | vedclass

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(N/A) Let $P(n): b_{n}=5+4n$ for all $n \in \mathbb{N}$.Step $1$: Base case for $n=1$.$P(1): b_{1}=5+4(1)=9$.From the recurrence relation $b_{k}=4+b_{

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(N/A) Let $P(n): b_{n}=5+4n$ for all $n \in \mathbb{N}$.
Step $1$: Base case for $n=1$.
$P(1): b_{1}=5+4(1)=9$.
From the recurrence relation $b_{k}=4+b_{k-1}$,for $k=1$,$b_{1}=4+b_{0}=4+5=9$.
Since $L.H.S. = R.H.S.$,$P(1)$ is true.
Step $2$: Inductive hypothesis.
Assume $P(k)$ is true for some $k \in \mathbb{N}$,i.e.,$b_{k}=5+4k$.
Step $3$: Inductive step.
We need to show $P(k+1)$ is true,i.e.,$b_{k+1}=5+4(k+1)$.
$b_{k+1}=4+b_{k}$ (by definition).
Substituting the hypothesis: $b_{k+1}=4+(5+4k) = 5+4(k+1)$.
Thus,$P(k+1)$ is true.
Conclusion: By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in \mathbb{N}$.

Use the Principle of Mathematical Induction to show that for a sequence $b_{0}, b_{1}, b_{2}, \ldots$ defined by $b_{0}=5$ and $b_{k}=4+b_{k-1}$ for all natural numbers $k$,the general term is $b_{n}=5+4n$ for all natural numbers $n$.

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