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Prove-by-mathematical-induction-that-2002-n-2-2003-2n-1-is-divisible-by-4005-




Question Number 85073 by Roland Mbunwe last updated on 18/Mar/20
Prove by mathematical induction that  2002^(n+2) +2003^(2n+1)     is divisible by 4005
$${Prove}\:{by}\:{mathematical}\:{induction}\:{that} \\ $$$$\mathrm{2002}^{{n}+\mathrm{2}} +\mathrm{2003}^{\mathrm{2}{n}+\mathrm{1}} \:\:\:\:{is}\:{divisible}\:{by}\:\mathrm{4005} \\ $$
Commented by MJS last updated on 18/Mar/20
it′s true for n=1 but it′s wrong for n=2, 3, 4
$$\mathrm{it}'\mathrm{s}\:\mathrm{true}\:\mathrm{for}\:{n}=\mathrm{1}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{wrong}\:\mathrm{for}\:{n}=\mathrm{2},\:\mathrm{3},\:\mathrm{4} \\ $$

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