A sequence `a_(1),a_(2),a_(3), . . .` is defined by letting `a_(1)=3` and `a_(k)=7a_(k-1)`, for all natural numbers `kge2`. Show that `a_(n)=3*7^(n-1)` for natural numbers.

Explanation

We have a sequence `a_(1),a_(2),a_(3)"…."` is defined by letting `a_(1) = 3` and `a_(k) = 7a_(k-1)`, for all natural number `k ge 2`.
Let `P(n) : a_(n) = 3 xx 7^(2-1) = 3 xx 7^(1) = 21`
Also, `a_(1) = 3, a_(k) = 7a_(k-1)`
`rArr a_(2) = 7a_(1) =a = 7 xx 3 = 21`
Thus, `P(2)` is true.
Now, assume that `P(k)` is true.
`:. a_(k)= 3 xx 7^(k-1)`
Now, to prove `P(k+1)` we have to show that
`a_(k+1) = 3 xx 7^(k+1-1)`
Given that `a_(k) = 7a_(k-1)`
So, `a_(k+1) = 7a_(k+1-1)`
`= 7a_(k)`
`= 7 xx 3 xx 7^(k-1)`
`= 3 xx 7^((k+1)-1)`
Hence, `P(k+1)` is true whenever `P(k)` is true.
So, by the principle of mathematical inducton, `P(n)` is true for any natural number `n`.