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If a(1)=1, a(2)=5 and a(n+2)=5a^(n+1)-6a...

If `a_(1)=1, a_(2)=5 and a_(n+2)=5a^(n+1)-6a_(n), n ge 1`. Show by using mathematical induction that `a_(n)=3^(n)-2^(n)`

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Let P(n)=`a_(n)=3^(n)-2^(n)`
When n=1, L.H.S.=`a_(1)=1` (given)
and R.H.S.=`3^(1)-2^(1)=3-2=1`
`therefore LHS=RHS`
When n=2, LHS=`a_(2)=5` (given)
and RHS=`3^(2)-2^(2)=5`
Thus LHS=RHS
Hence P(1) and P(2) are true……(A)
Let P(m) and P(m+1) be true
`Rightarrow {{:(,a_(m)=3^(m)-2^(m)),(,a_(m+1)=3^(m+1)-2^(m+1)):}.....(ii)`
To prove P(m+2) is true i.e. `a_(m+2)=3^(m+2)-2^(m+@)`
Given, `a_(n+2)=5a_(n+1)-6a_(n)`
`therefore a_(m+2)=5a_(m+1)-6a_(m)`
`=5(3^(m+1)-2^(m+1))-6(3^(m)-2^(m))`
`=(5.3^(m+1)-6.3^(m))-(5.2^(m+1)-6.2^(m))`
`3^(m)(15-6)-2^(m)(10-6)`
`=3^(m).9-2^(m)4`
`=3^(m+2)-2^(m+2)`
Hence P(m+2) is true whenever P(m) and P(m+1) are true ...(B)
From (A) and (B) by the principle of mathematical induction that P(n) is true for all natural number n.
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