3 2n when divided by 8 leaves the remainder 1

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We havePn: 32n leaves remainder of 1 when divided by 8.Step 1: Base CaseFor n=1, 32n=32=9R98=1, where R represents remainder.Hence Pn is true for n=1Step 2: Inductive hypothesisLet Pn be true for n=k i.e.32k leaves remainder of 1 when divided by 8 ;iR32k8=1Step 3: Inductive CaseConsider32k+1=32k.32=32k.9=32k8+1=8.32k+32kNow 8.32k is divisible by 8 whereas by statement i , we have , 32k leaves remainder of 1 when divided by 8.Hence  leaves a remainder of 1, when divided by 8.Hence Pn is true for n=k+1.Since Pn is true for n=1 and when Pk is true then Pk+1 is also true, hencePn is true for all nN

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