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.

A sequence `a_(1),a_(2),a_(3), . . .` is defined by letting `a_(1)=3`

1.	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.
Answer» 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)=37^(n-1)` for all natural numbers. Step I We observe P(2) is true. For n=2, `a_(2)=37^(2-1)=37^(1)=21` is true. As `a_(1)=3,a_(k)=7a_(k-1)` `rArra_(2)=7a_(2-1)=7a_(1)` `rArra_(2)7xx3=21 [becausea_(1)=3]` Step II Now, assume that P(n) is true for n=k. `P(k):a_(k)=37^(k-1)` Step III Now, to prove P(k+1) is true, we have to show that `P(k+1):a_(k+1)=37^(k+1-1)` `a_(k+1)=7a_(k+1-1)+7a_(k)` `=737^(k-1)=37^(k-+1)` So, P(k+1) is true, whenever p(k) is true. Hence, P(n) is true.