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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`. |
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Answer» 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`. |
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