A sequence `b_(0),b_(1),b_(2), . . .` is defined by letting `b_(0)=5` and `b_(k)=4+b_(k-1)`, for all natural number k. Show that `b_(n)=5+4n`, for all natural number n using mathematical induction.

Explanation

Consider the given statement,
`P(n):b_(n)=5+4n`, for natural numbers given that `b_(0)=5` and `b_(k)=4+b_(k-1)`
Step I P(1) is true
`P(1):b_(1)=5+4xx1=9`
As `b_(0)=5,b_(1)=4+b_(0)=4+5=0`
Hence, P(1) true.
Step II Now, assume that P(n) true for n=k.
`P(k):b_(k)=5+4k`
Step III Now, to prove P(k+1) is true, we have to show that
`:. P(k+1):b_(k=1)=5+4(k+1)`
`b_(k+1)=4+b_(k+1-1)`
`=4+b_(k)`
`=4+5+4k=5+4(k+1)`
So, by the mathematical induction P(k+1) is true whenever p(k) is ture, hence P(n) is true.