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From (2) we get: 7h 1 (7g+7)n+1 - 2-(7g+6) (2 - N-(79+6) II ig+1)k-(a+1)t-q +1 h=0 k=1 7h 1 79+7)n+q+1 - 2-(6q+6) = (S2q+1 – 2-(64+6) EII II-, (a+1)k-(q+1)t +1 h=0 k=1 So n 7h+1 1 2(79+7)n+q+2 - 2-(6q+5) = (2 - N-(74+6) L II T a+1)k-(q+1)t-q +1 h=0 k=1

From (2) we get: 7h 1 (7g+7)n+1 - 2-(7g+6) (2 - N-(79+6) II ig+1)k-(a+1)t-q +1 h=0 k=1 7h 1 79+7)n+q+1 - 2-(6q+6) = (S2q+1 – 2-(64+6) EII II-, (a+1)k-(q+1)t +1 h=0 k=1 So n 7h+1 1 2(79+7)n+q+2 - 2-(6q+5) = (2 - N-(74+6) L II T a+1)k-(q+1)t-q +1 h=0 k=1

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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(d) Subtracting 2m-(79+6) from both sides in (1), we have: 1 Im+1 Lm-(7q+6) (Pm-g – Pm-(8q+7), 1+ II-o Pm-(q+1)t-q t=0 and the following formula, for n > q + 1 Pa+1)n-(a+1)²–1)¯ P(a+1)n=[(a+2)²+(5q+3)] (b), n-(q+1) 1 = (L – N-(7q+6)II 1+II-o P(a+1)k-(a+1)t-q k=1 t=0 (2) P(a+1)n-(a+1)²–(a+1)) – 12(a+1)n-[(a+2)²+(49+3)] > (q+1)n n-(q+1) 1 = (Lq+1 – 2-(6q+6) II 1+ II-o P(a+1)k-(a+1)t k=1 From (2) we get: n 7h 1 Lra+Dnt1 - 2-(74+6) = (21 – 52–(7q+6)) L1I T Pla+1)k-(q+1)t-q + 1' h=0 k=1 t=0 n 7h 1 L(79+7)n+q+1 - 2-(6q+6) (Na+1 - 2-(64+6) 2I1T a+1)k-(a+1)t + 1 h=0 k=1 t=0 So 7h+1 1 2(79+7)n+g+2 (Ga+5) = (M1 – N-(74+6)) L 11 T Lio+1)k-(q+1)t-q +1 h=0 k=1 t=0 4
Transcribed Image Text:(d) Subtracting 2m-(79+6) from both sides in (1), we have: 1 Im+1 Lm-(7q+6) (Pm-g – Pm-(8q+7), 1+ II-o Pm-(q+1)t-q t=0 and the following formula, for n > q + 1 Pa+1)n-(a+1)²–1)¯ P(a+1)n=[(a+2)²+(5q+3)] (b), n-(q+1) 1 = (L – N-(7q+6)II 1+II-o P(a+1)k-(a+1)t-q k=1 t=0 (2) P(a+1)n-(a+1)²–(a+1)) – 12(a+1)n-[(a+2)²+(49+3)] > (q+1)n n-(q+1) 1 = (Lq+1 – 2-(6q+6) II 1+ II-o P(a+1)k-(a+1)t k=1 From (2) we get: n 7h 1 Lra+Dnt1 - 2-(74+6) = (21 – 52–(7q+6)) L1I T Pla+1)k-(q+1)t-q + 1' h=0 k=1 t=0 n 7h 1 L(79+7)n+q+1 - 2-(6q+6) (Na+1 - 2-(64+6) 2I1T a+1)k-(a+1)t + 1 h=0 k=1 t=0 So 7h+1 1 2(79+7)n+g+2 (Ga+5) = (M1 – N-(74+6)) L 11 T Lio+1)k-(q+1)t-q +1 h=0 k=1 t=0 4
In this work, we deal with the following nonlinear difference equation Im-(79+6) 1+ II-o Pm-(9+1)t-g Im+1 = 0, 1, .., (1) m = t%3D0 where N-(79+6), 2-(7g+5), · -1, No E (0, 0) is investigated. (d) We can generate the following formulas: L(79+7)n+r(q+1)+s+1 r-7)(4+1)+s+1 (1– (ITm-12-(mq+, 1+ (IIm=1 2- m-1)+s)/2(r-7)(a+1)+s+1 (mq+m-: n-1)+s) Lm3D n 7h+r ΣΠ IIm=1 P(a+1)t-(mq+m-1)+s + 1 1 h=0 k=1 (e) If 2(7g+7)n+(t–1)q+t → a(t-1)q+t 70 then 2(7g+7)n+6q+7 → 0 as n → o, If 2(79+7)n+tq+t → atg+t #0 then 2(7g+7)n+7q+7 → 0 as n → o. t = 1,6.
Transcribed Image Text:In this work, we deal with the following nonlinear difference equation Im-(79+6) 1+ II-o Pm-(9+1)t-g Im+1 = 0, 1, .., (1) m = t%3D0 where N-(79+6), 2-(7g+5), · -1, No E (0, 0) is investigated. (d) We can generate the following formulas: L(79+7)n+r(q+1)+s+1 r-7)(4+1)+s+1 (1– (ITm-12-(mq+, 1+ (IIm=1 2- m-1)+s)/2(r-7)(a+1)+s+1 (mq+m-: n-1)+s) Lm3D n 7h+r ΣΠ IIm=1 P(a+1)t-(mq+m-1)+s + 1 1 h=0 k=1 (e) If 2(7g+7)n+(t–1)q+t → a(t-1)q+t 70 then 2(7g+7)n+6q+7 → 0 as n → o, If 2(79+7)n+tq+t → atg+t #0 then 2(7g+7)n+7q+7 → 0 as n → o. t = 1,6.
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