n 8. Σn=1[5 (3) algebraically simplify 5 goes to D 22-1 5 29x7" 7n] 2^4x3 2₁n- 21" 3^.201 210 5.147 21" Im n=x (5.14" +70) - (1.6²+3) 2 X 2ㅜㅠ +21 • 2^. 24 • ½ 3", 2ny 7" - 2(6 un an +1 5 (14^+¹)-(6+¹) 21041 5 (14^") - \/(6^*) 2177¹ 5.14"-(6) 21n 5 (14" +14) - / ((+6) 21" +21 ( 56141) -(13-6^) = 2/1 21 (5499 210 5 (14^) - ½ (16") X 21n 5 (147) - 1/2(67) x. +70)-(+6+3) 1.47.111 = 3.190 = 3.19047419(. >1 Divergent
n 8. Σn=1[5 (3) algebraically simplify 5 goes to D 22-1 5 29x7" 7n] 2^4x3 2₁n- 21" 3^.201 210 5.147 21" Im n=x (5.14" +70) - (1.6²+3) 2 X 2ㅜㅠ +21 • 2^. 24 • ½ 3", 2ny 7" - 2(6 un an +1 5 (14^+¹)-(6+¹) 21041 5 (14^") - \/(6^*) 2177¹ 5.14"-(6) 21n 5 (14" +14) - / ((+6) 21" +21 ( 56141) -(13-6^) = 2/1 21 (5499 210 5 (14^) - ½ (16") X 21n 5 (147) - 1/2(67) x. +70)-(+6+3) 1.47.111 = 3.190 = 3.19047419(. >1 Divergent
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
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This is great, I looked at it for a long time. But can you explain how you simplified the equation where ( a sub n+1 divided by a sub 1) to 14^n (70-1/2*6*6/14^n) in the numerator? When I do it, I end up canceling out everything when I simplify. Please see the photo of my work and let me know where I went wrong... if you can. Thank you!
Transcribed Image Text:malu2116
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しの
(5-747) - (==64)
12
3
(E+₂97.7) (OL+ ubis) 7 =
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(19)²-(utiq
21n
(₁91) 7/- (uhl)s
210
(9+) = -(tl + util) {
21" +21
حاب
(₁+19) ²/ - (₁+u+1) G
21141
21771
(49) 7-utlog
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1 +432
(₁9) ² - ( +179 F
(5614") - (1.61)
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2
7"
27 +21
(5014" +70) - ( ½ •6² +3)x ZTA
tuz
19
utou?
97 - uhl • G
دنی
21"
utz
к либо ис
DEV
MI
21n
uL
-
2^4 x3n
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-
11
21n
util
3n
uL x uč
n
ст
6412064
Cichl
5
8. Σ=1[5(3)
algebraically
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