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`. 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 `kle2`. Show that `a_(n)=3*7^(n-1)` for natural numbers.
Answer» 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`.