Let `a_(0)=2,a_1=5` and for `n ge 2, a_n=5a_(n-1)-6a_(n-2)`, then prove by induction that `a_(n)=2^(n)+3^(n), forall n ge 0 , n in N`.

Explanation

Let `P(n) :a_(n)=2^(n)+3^(n),forall n ge 0, n in N`.
and `a_(0)=2,a_1=5` and for `nge 0, n in N`.
Step I For `n=0`,
`a_(0)=2^(0)+3^(0)=1+10=2`
which is true as `a_(0)=2`
Also , for `n=1,a_(1)=2^(1)+3^(1)=2+3=5`
which is also true as `a_1=5` .
Hence , `P(0) and P(1)` are true .
Step II Assume that `P(k-1) and P(k)` are true. Then , `a_(k-1)=2^(k-1)+3^(k-1)` .......(i)
where `a_(k-1)=5a_(k-2)-6a_(k-3) and a_(k)=2^(k)+3^(k)` ........(ii)
where `a_(k)=5a_(k-1)-6a_(k-2)`
Step III For `n=k+1`.
`P(k+1):a_(k+1)=2^(k+1)+3^(k+1), forall k ge 0, k in N`.
where `a_(k+1)=5a_(k)-6a_(k-1)`
Now, `a_(k+1)=5a_(k)-6a_(k-1)`
`=5(2^k+3^k)-6(2^(k-1)+3^(k-1))` [by using Eqs. (i) and (ii)]
`=5.2^(k)+5.3^(k)-6.2^(k-1)-6.3^(k-1)`
`=2^(k-1)(5.2-6)+3^(k-1)(5.3-6)`
`=2^(k-1) 4+3^(k-1).9=2^(k+1)+3^(k+1)`
`rArr a_(k+1)=2^(k+1)+3^(k+1)`
where `a_(k+1)=5a_(k)-6a_(k-1)`
This shows that the result is true for `n=k+1`. Hence by the second principle of mathematical induction , the result is true for `n in N, n ge 0`.