M-「-5-7 7 -5 uncover formulas because that the entries the Mn, wherein n is a confident integer. (Your formulas must not contain complicated numbers.) Mn =


You are watching: Find formulas for the entries of mn, where n is a positive integer.

Solved using basic knowledge of straight algebra

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M = -5-7> L7 - 5 allow a be the eigenvalue FM IM-XI1=0 -5 2 - 7 =0 T7 -5-1 -> (-5-7² +40=0 x ² + 10x + 74=0 d = -10 + V 100 - 296 --101 V-196 .-104148 2 0= -5 + 70 & 2 = -5-zi when 2,=-5+7i let 5c, = loci be its eigen vector (A-7,I) 5c, = 0 1-5-(-5+7) -7 -5-65+7)
Binomial expansion (a+b)n = nan-kbk - m1= -4 Galo i (-5+7in - i (-5-7iNTO Ti i LC-5+7in (-5-7in 1 21) L-1 is H i(-5+7)+i (-S-7R -(-5+7)+(-5570R 2i) <(-stzin-(-5-tin ns (-5+7)nti (-5-till take into consideration Mi (-5+72)n +i (-5-72)2 M = i)(-5+7:))+(-5-70) 07 M7 = 2 m (5) M-* (70) + 8 ne (sinhAik> Mini DC (-5)n-1 (78) * (1 + (-10%)> thus Mit zo i kis odd a. Once K is also wegeta K=2Ę KEINUŞoz
KE & S (-5) 1-2h 7 2K (w 2k (2) whese th> ecs the greatest integes role on ..Min - ns nc (-577-22 2h Dhal 22nc(-s) 92-9672 K=o -0 21 Kao
Consider MM 1/2 = -(-5+76)n + 65-72 = - ne (-5)n-k (70k + nG (5) B-k (-70k M.2 in ~ (-5)n-k (72) K (-1+(-k) - ② K = 0 therefore Mr =0 i kis even for once kisodd x = 2kt 1 we get M.2 = 7 nc (-530-2k-1 (7) 2K+ (i) 2K+ (-1+1-12K+1) 2K+L Kao 2kt | shaman (7) 2K+ 1Co 12k K =0 Info> nc =-21 Zkt! (-5)n-2k-1 (7) 2 kt (1) ho (-2) in (-5) = 2K-1 (ajakt ! Kao Ln n-2k-1 AC okt kul (2) (-1) K=a consider M7 = (-5+7in-(-5-7in =-(-(5+7in+(5-tun) = - Min .: Mn = 22 <2/2> in ~ (-5) 1-2 k-1 (7) 2 kit (,)k 2kt) Bro think about My = ns (-5+7inti (-5-tin. Min 28 ( 1) nc kG 5) "24 (-173 (G32k K- a


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- Ri mn M22 J 1/2 M Mn. 1/2> in n-26 (-5) (-DR 7 2K -2k-1 K (-1) 2h+ 7 2kt ko 69127 n-2k-1 0-2h -5) 27-2560572 - K2k 2k1 K=0