Menu Close

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-




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

Leave a Reply

Your email address will not be published. Required fields are marked *