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On principle of mathematical induction...

On principle of mathematical induction

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Let `p(n): A^(n) = [(1+2n,-4n),(n,1-2n)]`
step I for `n=1,`
LHS `=A^(1)=A`
and RHS `= [(1+2,-4),(1,1-2)]=[(3,-4),(1,-1)]`
rArr LHS=RHS
therefore, `p(1) is true.
step II Assume that `p(k) is true , then
` p(k): A^(k) = [(1+2K,-4K),(K,1-2K)]`
step III for n=k+1, we have to prove that
`p(k+1):A^(k+1)= [(3+2k,-4(k+1),(k+1,-1-2k)]`
` LHS = A^(k+1)=A^(k).A`
`[(1+2k,-4k),(k,-1-2k)][(3,-4),(1,-1)]`
`=[(3(1+2K)-4K,-4(1+2K)+4K),(3K+1(1-2K),-4K-1(1-2))]`
`=[(3+2k,-4(k+k1),(k+1,-1-2k)]= RHS`
therefore, p(k+1) is true.
Hence, by the principal of mathematical induction `p(n)` is true for all `n in N.`
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