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Question Number 90485 by Maclaurin Stickker last updated on 23/Apr/20

if x_0 =x_1 =1 and x_(n+1) =1996x_n +1997x_(n−1)   for n≥2. Find the remainder of  the division of x_(1996)  by 3

$${if}\:{x}_{\mathrm{0}} ={x}_{\mathrm{1}} =\mathrm{1}\:{and}\:{x}_{{n}+\mathrm{1}} =\mathrm{1996}{x}_{{n}} +\mathrm{1997}{x}_{{n}−\mathrm{1}} \\ $$$${for}\:{n}\geqslant\mathrm{2}.\:{Find}\:{the}\:{remainder}\:{of} \\ $$$${the}\:{division}\:{of}\:{x}_{\mathrm{1996}} \:{by}\:\mathrm{3} \\ $$

Commented by Maclaurin Stickker last updated on 23/Apr/20

I find the general term of sequence  x_n =((998(−1)^n +1997^n )/(999))  If it is right, how could I find the remainder  without Newton′s Binomial?

$${I}\:{find}\:{the}\:{general}\:{term}\:{of}\:{sequence} \\ $$$${x}_{{n}} =\frac{\mathrm{998}\left(−\mathrm{1}\right)^{{n}} +\mathrm{1997}^{{n}} }{\mathrm{999}} \\ $$$${If}\:{it}\:{is}\:{right},\:{how}\:{could}\:{I}\:{find}\:{the}\:{remainder} \\ $$$${without}\:{Newton}'{s}\:{Binomial}? \\ $$

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