Given that `u_(n+1)=3u_n-2u_(n-1),` and `u_0=2 ,u_(1)=3`, then prove that `u_n=2^(n)+1` for all positive integer of `n`

Explanation

`because `U_(n+1)=3U_(n)-2U_(n-1)`
Step I `U_(1)=3=2+1=2^1+1` which is true for `n=1`.
Putting `n=1` in Eq. (i) we get
`U_(1+1)=3U_(1)-2U_(1-1)`
`rArr U_(2)=3u_(1)-2u_(0)=3.3-2.2=5=2^2+1` which is true for `n=2`
Therefore , the result is true for `n =1` and n=2`.
Step II Assume it is true for `n=k` , then it is also true for `n=k-`.
Then , `U_(k)=2^(k)+1`.....(ii)
and `u_(k-1)=2^(k-1)+1` ......(iii)
Step III Putting `n=k` in Eq. (i) we get
u_(k+1)=3u_(k)-2u_(k-1)`
`=3(2^k+1)-2(2^k-1+1)` [from Eqs. (ii) and (iii)]
`3.2^k+3-2.2^k-1-2=3.2^k+3-2^k-2`
`=(3-1)2^k+1=2.2^k+1=2^k+1+1`
This shows that the resutl is true for `n=k+1`. Hence , by the principle of mathematical induction the result is true for all `n in N`.