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When the square of any odd number, greater than1, is divided by 8, it always leaves remainder1(b) 6 (c) 8 (d) Cannot be determined |
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Answer» Let `P(n):(2r+1)^(2n), forall n in N and r in I`. Step I For `n=1`. `P(1):(2r+1)^2=4r^2+4r+1=4r(r+1)+1=8p+1,p in I " "[because r(r+1)"is an even integer"]` Therefore , `P(1)` is true , Step II Assume P(n) is true for n=k , then `P(k):(2r+1)^2k` is divisible by 8 levaes remainder 1. `rArr P(k)=8m+1,n in I`, where m is a positive integer . Step III For `n=k+1`. brgt `therefore P(k)=(2r+1)2(k+1)` `=(2r+1)^(2k)(2r+1)^2` `=(8m+1)(8p+1)` `64mp+8(m+p)+1` `=8(8mp+m+p)+1` which is true for `n=k+1` as `8mp+m+p` is an integer. Hence , by the principle of mathematical induction, when P(n) is divided by 8 leaves the ramainder 1 for all `n in N`. |
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