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 `k≥2`. Show that `a_(n)=3*7^(n-1)` for natural numbers.

Explanation

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 `kle2`.
Let `P(n):a_(n)=3*7^(n-1)` for all natural numbers.
Step I We observe P(2) is true.
For n=2, `a_(2)=3*7^(2-1)=3*7^(1)=21` is true.
As `a_(1)=3,a_(k)=7a_(k-1)`
`rArra_(2)=7*a_(2-1)=7*a_(1)`
`rArra_(2)7xx3=21 [becausea_(1)=3]`
Step II Now, assume that P(n) is true for n=k.
`P(k):a_(k)=3*7^(k-1)`
Step III Now, to prove P(k+1) is true, we have to show that
`P(k+1):a_(k+1)=3*7^(k+1-1)`
`a_(k+1)=7*a_(k+1-1)+7*a_(k)`
`=7*3*7^(k-1)=3*7^(k-+1)`
So, P(k+1) is true, whenever p(k) is true. Hence, P(n) is true.