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2. n-1 X6n-2 = r II i=0 1/^) (fot foi+2p+f6i+1h) f6i+1P+ foih f6i+49 + foi+3k f6i+39 +f6i+2k)

2. n-1 X6n-2 = r II i=0 1/^) (fot foi+2p+f6i+1h) f6i+1P+ foih f6i+49 + foi+3k f6i+39 +f6i+2k)

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter2: Parallel Lines
Section2.CT: Test
Problem 3CT: To prove a theorem of the form "If P, then Q" by the indirect method, the first line of the proof...
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Bxn-10n-2 YIn-1 + dxn-4 Xn+1 = axn-2 + n = 0,1, ..., (1) The following special case of Eq.(1) has been studied Xn-1Xn-2 In+1 = Tn-2 + (8) In-1+ Xn-4' where the initial conditions x-4, x-3, x-2, -1,and xo are arbitrary non zero real numbers. Theorem 4. Let {Tn}-4 be a solution of Eq.(8). Then for n = 0, 1, 2,..
Transcribed Image Text:Bxn-10n-2 YIn-1 + dxn-4 Xn+1 = axn-2 + n = 0,1, ..., (1) The following special case of Eq.(1) has been studied Xn-1Xn-2 In+1 = Tn-2 + (8) In-1+ Xn-4' where the initial conditions x-4, x-3, x-2, -1,and xo are arbitrary non zero real numbers. Theorem 4. Let {Tn}-4 be a solution of Eq.(8). Then for n = 0, 1, 2,..
r(+ T (fait-sP + fes+7h) (feira + feits + Son +24 + Son+zki] п-2 2p + h П fei+sP+ fei+7h foi+7P+ foi+sh ) \ Foi+39 + foi+2k / ( foi+49 + fei+3k` X6n-2 p+h i=0 п-2 2p+h p+h f6i+49+f6i+3k f6i+39+f6i+2k h r f6i+7p+f6i+6h i=0 ( fönq+fên–1k ) h fön+19+fénk h +P п-2 foi+8P + f6i+7h` П föi+TP + foi+6h, ( foi+49 + f6i+3k` föi+39 + f6i+2k, 2p + h X6n-2 p+h i=0 n-2 2p+h p+h П f6i+8p+f6i+7h f6i+7p+f6i+6h f6i+49+ƒ6i+3k f6i+39+f6i+2k h r i=0 h1+ fénq+ fon-1k fén+19+f6nk n-2 2p + h П foi+8P+ f6i+7h foi+7P+ foi+sh ) \foi+34 + foi+2k , ( foi+49 + fei+3k` X6n-2 p+h i=0 n-2 2p+h |Tei4Tp+f6i+6h p+h foi+8p+f6i+7h f6i+49+f6i+3k f6i+39+f6i+2k i=0 +- fên+19+fénk+fênq+fén–1k fön+19+f6nk п-2 2р + h r( IIa-p+ fansh) \fei-39 + fei+zk ) (föi+49 + fsi+3k` p+h i=0 п-2 2p+h p+h IT ( föi+8p+f6i+7h f6i+7p+f6i+6h f6i+49+ƒ6i+3k f6i+3q+f6i+2k i=0 fon+29+f6n+1k fön+19+f6nk п-2 (2р + h П föi+7P+ f6i+6h, föi+49 + f6i+3k foi+39 + fói+2k ) fon+19 + fönk 1+ p+h fön+24 + fön+1k ] i=0 п-2 (2p +h П foi+7P + foi+6h, foi+8P+ foi+7h` fei+49 + fei+3k\ [fon+29 + fén+1k + fên+19 + f&nk¯ foi+39 + foi+2k, p+h fön+24 + fön+ik i=0 п-2 fei+sp+ fei+7h` П foi+7p + foi+ch ) \ foi+39 + föi+2k ) [fon+29 + fön+1k 2p + h (fei+4q + fei+3k\ [fön+39 + fen+2k] = r p+h i=0 16 Therefore foi+2P+ föi+1h П foi+1p+ fesh )( oi+49 + foi+3k п-1 X6n-2 = r fei+39 + foi+2k, i=0 Also, other formulas can be proved similarly. Hence, the proof is completed.
Transcribed Image Text:r(+ T (fait-sP + fes+7h) (feira + feits + Son +24 + Son+zki] п-2 2p + h П fei+sP+ fei+7h foi+7P+ foi+sh ) \ Foi+39 + foi+2k / ( foi+49 + fei+3k` X6n-2 p+h i=0 п-2 2p+h p+h f6i+49+f6i+3k f6i+39+f6i+2k h r f6i+7p+f6i+6h i=0 ( fönq+fên–1k ) h fön+19+fénk h +P п-2 foi+8P + f6i+7h` П föi+TP + foi+6h, ( foi+49 + f6i+3k` föi+39 + f6i+2k, 2p + h X6n-2 p+h i=0 n-2 2p+h p+h П f6i+8p+f6i+7h f6i+7p+f6i+6h f6i+49+ƒ6i+3k f6i+39+f6i+2k h r i=0 h1+ fénq+ fon-1k fén+19+f6nk n-2 2p + h П foi+8P+ f6i+7h foi+7P+ foi+sh ) \foi+34 + foi+2k , ( foi+49 + fei+3k` X6n-2 p+h i=0 n-2 2p+h |Tei4Tp+f6i+6h p+h foi+8p+f6i+7h f6i+49+f6i+3k f6i+39+f6i+2k i=0 +- fên+19+fénk+fênq+fén–1k fön+19+f6nk п-2 2р + h r( IIa-p+ fansh) \fei-39 + fei+zk ) (föi+49 + fsi+3k` p+h i=0 п-2 2p+h p+h IT ( föi+8p+f6i+7h f6i+7p+f6i+6h f6i+49+ƒ6i+3k f6i+3q+f6i+2k i=0 fon+29+f6n+1k fön+19+f6nk п-2 (2р + h П föi+7P+ f6i+6h, föi+49 + f6i+3k foi+39 + fói+2k ) fon+19 + fönk 1+ p+h fön+24 + fön+1k ] i=0 п-2 (2p +h П foi+7P + foi+6h, foi+8P+ foi+7h` fei+49 + fei+3k\ [fon+29 + fén+1k + fên+19 + f&nk¯ foi+39 + foi+2k, p+h fön+24 + fön+ik i=0 п-2 fei+sp+ fei+7h` П foi+7p + foi+ch ) \ foi+39 + föi+2k ) [fon+29 + fön+1k 2p + h (fei+4q + fei+3k\ [fön+39 + fen+2k] = r p+h i=0 16 Therefore foi+2P+ föi+1h П foi+1p+ fesh )( oi+49 + foi+3k п-1 X6n-2 = r fei+39 + foi+2k, i=0 Also, other formulas can be proved similarly. Hence, the proof is completed.
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