n-2 (2p + h ( fi+8P+ fei+7h` fei+7P + fsi+sh ) (föi+39 + fei+2k ) ( foi+49 + fei+3k° fon+19 + fonk fon+29 + fon+1k] p+h n-2 (2p + h` p+h (fei+8P+ fei+7h` Soi+7P + foi+sh, Soi+39 + fessak ) |n+29 + Jon+1k + fon+19 + fôn+29 + fón+1k foi+49 + fei+3k` r (2p + h) ÏT ) (fei+49 + fei+3k\ ( foi+8P + fei+7h` foi+49 + fei+3k` (Joi+7P+ fei+sh ) \fei+39 + fei+zk ) [fon+29 + fon+1k] [ fon+39 + fon+2k] p+h i=0 16 Therefore n- rT foi-2p+ fei+1h fei+1p+ feih ) (oi+49 + fei+3k` I Iôn-2 foi+39 + fei+2k, i=0
n-2 (2p + h ( fi+8P+ fei+7h` fei+7P + fsi+sh ) (föi+39 + fei+2k ) ( foi+49 + fei+3k° fon+19 + fonk fon+29 + fon+1k] p+h n-2 (2p + h` p+h (fei+8P+ fei+7h` Soi+7P + foi+sh, Soi+39 + fessak ) |n+29 + Jon+1k + fon+19 + fôn+29 + fón+1k foi+49 + fei+3k` r (2p + h) ÏT ) (fei+49 + fei+3k\ ( foi+8P + fei+7h` foi+49 + fei+3k` (Joi+7P+ fei+sh ) \fei+39 + fei+zk ) [fon+29 + fon+1k] [ fon+39 + fon+2k] p+h i=0 16 Therefore n- rT foi-2p+ fei+1h fei+1p+ feih ) (oi+49 + fei+3k` I Iôn-2 foi+39 + fei+2k, i=0
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,...
Related questions
Question
Show me the steps of determine red and information is here step by step .it complete
![n-2
[Son-3P+ fön-4h + fên-4P+ fón-5h
fön-3p + fén-gh
(Söi+aP + fei+3h\
Soi+69 + fei+sk
II
fei+3P+ fei+2h ) (foi+59 + fei+ak )
i=0
n-2
fei+4P + foi+3h (foir69 + fei+sk)
(fei+3p + f6i+2h ) föi+59 + fei+ak ) (fên-3p + fön-4h
fon-2P + fén-3h
= q
i=0
Therefore
Xôn-3 = kI( I6i+4P + fei+ah
foi+3P + fei+2h )
n-1
feiq + fei-1k
foi-19 + fei-2k,
i=0
Also, from Eq.(8), we see that
X6n-4X6n-5
I6n-2 = X6n-5 +
X6n-4 + xôn-7
n-2
(2p + h
+h) IIFoi+7p+ fsi+sh ) \ foi+39 + fei+2k ,
foi+sp+ fei+7h`
( foit49 + föi+3k`
i=0
n-1
n-2
Jei+39+J6i+2k
+i=0
n-1
i=0
n-2
Jei+5P+/ei+4h
S6i+19+Seik
i=0
n-2
(2р + h)
fei+7P+ f6i+6h) (föi+39 + fei+2k )
i=0
n-2
+
n-2
Jen+19+ Jen
Jeng+fen-1k
i=0
i=0
i=0
n-2
fei+8P + fei+7h\ ( fei+49 + fei+3k`
fei+7p + fei+oh)\ fei+34 + f6i+2k,
(2p + h \
)II
= r
p+h
i=0
n-2
(2p+h
pth
(Tei+7p+Sei+ah
i=0
fei+49+S6i+ak\
Jei+39+Se+zk)
+
n-1
n-2
fei-1p+ fei-2h
fei46p+ fei4sh
(Leng+fen-1k
h+P
Jen+19+fonk
f6i+sp+ fei+ah
| fsipt Sei-ih
i=0
i=0
15
- -(
п-2
föi+49 + fei+3k \
foi+sP + fei+7h\
( fei+7P + f6i+oh )\ fôi+39 + fei+2k)
2р +
p+h)I
I6n-2
i=0
T-2
(2p+h) T (fsitsp+fai47h
S6i+49+So143k
Joi+39+ Sei+2k)
h
i=0
+
fonq+fon-ik
( Jên+19+fenk)
h + p
n-2
(2р + h)
p+h
I6n-2
fei+7P + fei+6h
( Soi+39 + f6i+2k )
i=0
n-2
hr
feit49+Sei+3k
i=0
+
h
Jên +19+ fenk
n-2
2р +
föi+49 + fei+3k`
6n-2
= r
p+h
Sei+7P + fei+sh ) \ Foi+39 + fei+2k )
i=0
n-2
Joi+7p+fes+6h) (Tei+39+Sei+2k
i=0
fon +19+fank+Sanq+Son-1k
Sen+19+ fenk
n-2
( foi+8P + f6i+7h`
foi+7P + fei+ch )foi+39 + foi+2k,
2p + h
föi+49 + foi+3k\
=
P+h
i=0
n-2
2p+h
p+h
Joi+7p+fei+sh) Foi439+fei+2k
i=0
fen+29+føn+1k
fon+19+fenk
n-2
fei+49 + fei+3k`
Joi+7P + foi+oh ) (Foi-39 + fei+2k ) [* + fon+29 + fon+1k]
'2р + h)
fön+19 + fenk
p+h
i=0
(2p+ h T (foit8P+ fei+7h\ (fei+49 + fei+3k\ [fon+29 + fön+1k + fon+19+ fonk"
foi+7P + foi+6h) (foi+39 + foi+2k /
п-2
fon+29 + fön+1k
i=0
п-2
fei+4q + fei+3k\ [ fon+39 + f6n+2k]
föi+7P + fei+6h ) (fBi+39 + f6i+2k) [fön+29 + fén+1k
(2p + h)
AP+h
i=0
16
Therefore
( föi+2P + fei+1h`
fei+ip + fesh)(Gi+49 + foi+3k*
n-1
X6n-2 = r
foi+39 + f6i+2k
i=0
Also, other formulas can be proved similarly. Hence, the proof is completed.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F158aed3a-44ed-4147-bb6c-9729d6a7101b%2F67438456-d713-4bb9-9745-39781b7a8973%2Fn05biz_processed.png&w=3840&q=75)
Transcribed Image Text:n-2
[Son-3P+ fön-4h + fên-4P+ fón-5h
fön-3p + fén-gh
(Söi+aP + fei+3h\
Soi+69 + fei+sk
II
fei+3P+ fei+2h ) (foi+59 + fei+ak )
i=0
n-2
fei+4P + foi+3h (foir69 + fei+sk)
(fei+3p + f6i+2h ) föi+59 + fei+ak ) (fên-3p + fön-4h
fon-2P + fén-3h
= q
i=0
Therefore
Xôn-3 = kI( I6i+4P + fei+ah
foi+3P + fei+2h )
n-1
feiq + fei-1k
foi-19 + fei-2k,
i=0
Also, from Eq.(8), we see that
X6n-4X6n-5
I6n-2 = X6n-5 +
X6n-4 + xôn-7
n-2
(2p + h
+h) IIFoi+7p+ fsi+sh ) \ foi+39 + fei+2k ,
foi+sp+ fei+7h`
( foit49 + föi+3k`
i=0
n-1
n-2
Jei+39+J6i+2k
+i=0
n-1
i=0
n-2
Jei+5P+/ei+4h
S6i+19+Seik
i=0
n-2
(2р + h)
fei+7P+ f6i+6h) (föi+39 + fei+2k )
i=0
n-2
+
n-2
Jen+19+ Jen
Jeng+fen-1k
i=0
i=0
i=0
n-2
fei+8P + fei+7h\ ( fei+49 + fei+3k`
fei+7p + fei+oh)\ fei+34 + f6i+2k,
(2p + h \
)II
= r
p+h
i=0
n-2
(2p+h
pth
(Tei+7p+Sei+ah
i=0
fei+49+S6i+ak\
Jei+39+Se+zk)
+
n-1
n-2
fei-1p+ fei-2h
fei46p+ fei4sh
(Leng+fen-1k
h+P
Jen+19+fonk
f6i+sp+ fei+ah
| fsipt Sei-ih
i=0
i=0
15
- -(
п-2
föi+49 + fei+3k \
foi+sP + fei+7h\
( fei+7P + f6i+oh )\ fôi+39 + fei+2k)
2р +
p+h)I
I6n-2
i=0
T-2
(2p+h) T (fsitsp+fai47h
S6i+49+So143k
Joi+39+ Sei+2k)
h
i=0
+
fonq+fon-ik
( Jên+19+fenk)
h + p
n-2
(2р + h)
p+h
I6n-2
fei+7P + fei+6h
( Soi+39 + f6i+2k )
i=0
n-2
hr
feit49+Sei+3k
i=0
+
h
Jên +19+ fenk
n-2
2р +
föi+49 + fei+3k`
6n-2
= r
p+h
Sei+7P + fei+sh ) \ Foi+39 + fei+2k )
i=0
n-2
Joi+7p+fes+6h) (Tei+39+Sei+2k
i=0
fon +19+fank+Sanq+Son-1k
Sen+19+ fenk
n-2
( foi+8P + f6i+7h`
foi+7P + fei+ch )foi+39 + foi+2k,
2p + h
föi+49 + foi+3k\
=
P+h
i=0
n-2
2p+h
p+h
Joi+7p+fei+sh) Foi439+fei+2k
i=0
fen+29+føn+1k
fon+19+fenk
n-2
fei+49 + fei+3k`
Joi+7P + foi+oh ) (Foi-39 + fei+2k ) [* + fon+29 + fon+1k]
'2р + h)
fön+19 + fenk
p+h
i=0
(2p+ h T (foit8P+ fei+7h\ (fei+49 + fei+3k\ [fon+29 + fön+1k + fon+19+ fonk"
foi+7P + foi+6h) (foi+39 + foi+2k /
п-2
fon+29 + fön+1k
i=0
п-2
fei+4q + fei+3k\ [ fon+39 + f6n+2k]
föi+7P + fei+6h ) (fBi+39 + f6i+2k) [fön+29 + fén+1k
(2p + h)
AP+h
i=0
16
Therefore
( föi+2P + fei+1h`
fei+ip + fesh)(Gi+49 + foi+3k*
n-1
X6n-2 = r
foi+39 + f6i+2k
i=0
Also, other formulas can be proved similarly. Hence, the proof is completed.
![Brn-1In-2
YIn-1 + &xn-4
In+1 = a1n-2 +
n = 0, 1,...,
(1)
1
The following special case of Eq.(1) has been studied
In-1In-2
Int1 = In-2+
(8)
Tn-1 + In-4
where the initial conditions r-4, I-3, T-2, T-1,and xo are arbitrary non zero real
numbers.
Theorem 4. Let {zn}-4 be a solution of Eq.-(8). Then for n = 0, 1,2, ...
fep + fes-ih ( foi+29 + foiik
Jei-1p+ fei-zh)
fo419 + Sauk)
farq + Sai-1k
feirap + forah ) To-19 + foi-ak ,
(Sei+aP + fot+3h)
fei-2P + fei+1h (foi+19 + Sei+ak
= rT
fei+1P+ fesh ) Joi+39 + fei+2k )
(fei+ep + fes+sh\ (foi+29 + fei+ik
Soi-sp + feisah)
fei+ap + fei+ah\ ( feiseq + fei-sk
Sai+ap + fei+2h ) Jai459 + Seisak )
fei+19 + fask
2p +h
II ap+ fesch) (Tars9 + Suvzk,
foi+sP + fouth ( Seisa9 + Soirak
In+1 =
p+h
where r-4 = h, r-3 = k, z-2 = r, r-1 = P, 2o = q, {fm}-1 = {1,0,1,1,2, 3, 5, 8, .}.
Proof: For n -0 the result holds. Now suppose that n>0 and that our assumption
holds for n - 2. That is;
feirap + foi+ah
Jei+ap + fei42h) J6i-19 + fei-2k
fauq + foi-1 k
i-0
fai+1p + fesh ) fet439 + feirak)
Joi+ep + fei+sh (fei+29 + foi+zk
foi+sp + fes+ah)
Iom-6 = 9TT(fei-4p + foi+3h (fei+09 + fei-sk
+ fei+zh) \Torr39 + fei+ak ) *
(2p + h ( foi-sp+ fesuzh (foir49 + foi+ak
Ji+7p + Soisgh) Tausg + Soi+zk
p+h
Now, it follows from Eq.(8) that
IGn-4 = Iộn-7 +
11
feisep + fo+sh ( foi+29 + fe+ik
Sunsp + Seissh ) (Tus1g+ fesk)
= p||
•II )( )II( ) )
(foi+op + fou-sh ( fei-29 + fei+ikY
1foi+sp + foi+sh)
foi+19 + fosk
+ A
PIT fasop + foish) ( for-29 + Sousik
Joi4sp+fostah)
n-2
= p
Fei+sp + feirah)
feir19 + faik
pg
( foi+6P + fei+sh) ( fei+29 + fei-ik
=PlI(asp+ fash)( Far1q + Sak )
12
PII
´ fouvap + faunh\ ( fa-29 + feieik)
Sos+19 + Souk
PIIusp + Jourah)
1+ (
PII( ) )
-PT (furep + fouosh (fau29 + fuosk
Sei-sp + foi-gh )
fois19 + fuk)
n-2
PII( (unt
Jeusp + Sanah)
Sesig+ fauk
(An
( fesep + fursh ( fev29 + Jairak1+a+ fon-ek
- PII
Sen-o9 + fon-zk]
n-2
Sen-s9 + Son -ak-"
Sei+ep + Sei+sh) ( Su-29 + Sai+ik [ Sen-sq + Sen-sk
= PII(sp+ fursh)Tuer9+ feck ) [Ton-39 + Sen-sk]
Therefore
R-1
Also, from Eq.(8), we see that
Tusap + Seurzli ) ( Jonnog + Senek )
( )--II( )( )
13](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F158aed3a-44ed-4147-bb6c-9729d6a7101b%2F67438456-d713-4bb9-9745-39781b7a8973%2Fftemkhq_processed.png&w=3840&q=75)
Transcribed Image Text:Brn-1In-2
YIn-1 + &xn-4
In+1 = a1n-2 +
n = 0, 1,...,
(1)
1
The following special case of Eq.(1) has been studied
In-1In-2
Int1 = In-2+
(8)
Tn-1 + In-4
where the initial conditions r-4, I-3, T-2, T-1,and xo are arbitrary non zero real
numbers.
Theorem 4. Let {zn}-4 be a solution of Eq.-(8). Then for n = 0, 1,2, ...
fep + fes-ih ( foi+29 + foiik
Jei-1p+ fei-zh)
fo419 + Sauk)
farq + Sai-1k
feirap + forah ) To-19 + foi-ak ,
(Sei+aP + fot+3h)
fei-2P + fei+1h (foi+19 + Sei+ak
= rT
fei+1P+ fesh ) Joi+39 + fei+2k )
(fei+ep + fes+sh\ (foi+29 + fei+ik
Soi-sp + feisah)
fei+ap + fei+ah\ ( feiseq + fei-sk
Sai+ap + fei+2h ) Jai459 + Seisak )
fei+19 + fask
2p +h
II ap+ fesch) (Tars9 + Suvzk,
foi+sP + fouth ( Seisa9 + Soirak
In+1 =
p+h
where r-4 = h, r-3 = k, z-2 = r, r-1 = P, 2o = q, {fm}-1 = {1,0,1,1,2, 3, 5, 8, .}.
Proof: For n -0 the result holds. Now suppose that n>0 and that our assumption
holds for n - 2. That is;
feirap + foi+ah
Jei+ap + fei42h) J6i-19 + fei-2k
fauq + foi-1 k
i-0
fai+1p + fesh ) fet439 + feirak)
Joi+ep + fei+sh (fei+29 + foi+zk
foi+sp + fes+ah)
Iom-6 = 9TT(fei-4p + foi+3h (fei+09 + fei-sk
+ fei+zh) \Torr39 + fei+ak ) *
(2p + h ( foi-sp+ fesuzh (foir49 + foi+ak
Ji+7p + Soisgh) Tausg + Soi+zk
p+h
Now, it follows from Eq.(8) that
IGn-4 = Iộn-7 +
11
feisep + fo+sh ( foi+29 + fe+ik
Sunsp + Seissh ) (Tus1g+ fesk)
= p||
•II )( )II( ) )
(foi+op + fou-sh ( fei-29 + fei+ikY
1foi+sp + foi+sh)
foi+19 + fosk
+ A
PIT fasop + foish) ( for-29 + Sousik
Joi4sp+fostah)
n-2
= p
Fei+sp + feirah)
feir19 + faik
pg
( foi+6P + fei+sh) ( fei+29 + fei-ik
=PlI(asp+ fash)( Far1q + Sak )
12
PII
´ fouvap + faunh\ ( fa-29 + feieik)
Sos+19 + Souk
PIIusp + Jourah)
1+ (
PII( ) )
-PT (furep + fouosh (fau29 + fuosk
Sei-sp + foi-gh )
fois19 + fuk)
n-2
PII( (unt
Jeusp + Sanah)
Sesig+ fauk
(An
( fesep + fursh ( fev29 + Jairak1+a+ fon-ek
- PII
Sen-o9 + fon-zk]
n-2
Sen-s9 + Son -ak-"
Sei+ep + Sei+sh) ( Su-29 + Sai+ik [ Sen-sq + Sen-sk
= PII(sp+ fursh)Tuer9+ feck ) [Ton-39 + Sen-sk]
Therefore
R-1
Also, from Eq.(8), we see that
Tusap + Seurzli ) ( Jonnog + Senek )
( )--II( )( )
13
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