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Question Number 139745 by mathocean1 last updated on 01/May/21
n ∈ N.  a_n =5^(n+1) +5^n  We can say that:  • a_n  is always odd.  • a_n  is always even.  • It depends of n.  • The sign of a_n  depends of n.
$${n}\:\in\:\mathbb{N}. \\ $$$${a}_{{n}} =\mathrm{5}^{{n}+\mathrm{1}} +\mathrm{5}^{{n}} \:{We}\:{can}\:{say}\:{that}: \\ $$$$\bullet\:{a}_{{n}} \:{is}\:{always}\:{odd}. \\ $$$$\bullet\:{a}_{{n}} \:{is}\:{always}\:{even}. \\ $$$$\bullet\:{It}\:{depends}\:{of}\:{n}. \\ $$$$\bullet\:{The}\:{sign}\:{of}\:{a}_{{n}} \:{depends}\:{of}\:{n}. \\ $$$$ \\ $$$$ \\ $$
Answered by floor(10²Eta[1]) last updated on 01/May/21
a_n =5^(n+1) +5^n =5^n (5+1)=5^n 6   ⇒a_n  is always even
$$\mathrm{a}_{\mathrm{n}} =\mathrm{5}^{\mathrm{n}+\mathrm{1}} +\mathrm{5}^{\mathrm{n}} =\mathrm{5}^{\mathrm{n}} \left(\mathrm{5}+\mathrm{1}\right)=\mathrm{5}^{\mathrm{n}} \mathrm{6}\: \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{n}} \:\mathrm{is}\:\mathrm{always}\:\mathrm{even} \\ $$
Commented by mathocean1 last updated on 01/May/21
thanks
$${thanks} \\ $$

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