For any n in N,4^n+15n-1 is divisible by | vedclass

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(B) Let $P(n): 4^n+15n-1$ be divisible by $9$. For $n=1$,$P(1)=4^1+15(1)-1=18$,which is divisible by $9$. Therefore,$P(1)$ is true. Assume $P(k)$ is t

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$9$

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(B) Let $P(n): 4^n+15n-1$ be divisible by $9$. For $n=1$,$P(1)=4^1+15(1)-1=18$,which is divisible by $9$. Therefore,$P(1)$ is true. Assume $P(k)$ is true for some $k \in N$. Then,$4^k+15k-1=9\lambda$ for some $\lambda \in N$. We need to show that $P(k+1)$ is true,i.e.,$4^{k+1}+15(k+1)-1$ is divisible by $9$. $4^{k+1}+15(k+1)-1 = 4 \cdot 4^k + 15k + 15 - 1$ Substitute $4^k = 9\lambda - 15k + 1$: $= 4(9\lambda - 15k + 1) + 15k + 14$ $= 36\lambda - 60k + 4 + 15k + 14$ $= 36\lambda - 45k + 18$ $= 9(4\lambda - 5k + 2)$,which is divisible by $9$. Thus,$P(k+1)$ is true whenever $P(k)$ is true. By the Principle of Mathematical Induction,$4^n+15n-1$ is divisible by $9$ for all $n \in N$.

For any $n \in N$,$4^n+15n-1$ is divisible by

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