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A sequence $a_1, a_2, a_3, \ldots$ is defined by letting $a_1=3$ and $a_k=7 a_{k-1}$, for all natural numbers $k \geq 2$. Show that $a_n=3^n \cdot \overline{7}^{n-1}$ for all natural numbers,
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Let $P(n): a_n=3 \cdot 7^{n-1}$ for all natural numbers. For $n=2$,
$a_2=3 \cdot 7^{2-1}=3 \cdot 7^1=21$ is true.
As $\quad a_1=3, a_k=7 a_{k-1}$
$\Rightarrow \quad a_2=7 \cdot a_{2-1}=7 \cdot a_1$
$\Rightarrow \quad a_2=7 \times 3=21 \quad\left[\because a_1=3\right]$
Now, let $P(n)$ is true for $n=k$.
$$
P(k): a_k=3 \cdot 7^{k-1}
$$
Now, to prove $P(k+1)$ is true, we have to show that $P(k+1): a_{k+1}=3 \cdot 7^{k+1-1}$
Now, $\quad a_{k+1}=7 \cdot a_{k+1-1}=7 \cdot a_k$
$$
=7 \cdot 3 \cdot 7^{k-1}=3 \cdot 7^{k-1+1}
$$
So, $P(k+1)$ is true, whenever $p(k)$ is true. Hence, $P(n)$ is true.
$a_2=3 \cdot 7^{2-1}=3 \cdot 7^1=21$ is true.
As $\quad a_1=3, a_k=7 a_{k-1}$
$\Rightarrow \quad a_2=7 \cdot a_{2-1}=7 \cdot a_1$
$\Rightarrow \quad a_2=7 \times 3=21 \quad\left[\because a_1=3\right]$
Now, let $P(n)$ is true for $n=k$.
$$
P(k): a_k=3 \cdot 7^{k-1}
$$
Now, to prove $P(k+1)$ is true, we have to show that $P(k+1): a_{k+1}=3 \cdot 7^{k+1-1}$
Now, $\quad a_{k+1}=7 \cdot a_{k+1-1}=7 \cdot a_k$
$$
=7 \cdot 3 \cdot 7^{k-1}=3 \cdot 7^{k-1+1}
$$
So, $P(k+1)$ is true, whenever $p(k)$ is true. Hence, $P(n)$ is true.
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