By shifting the summation index of the following added three series Σ n(n-1)₂x + [n n(n-1)C₂x-²-4n1"+ WW2 192 we get: Σ[(k-2)(k-3)Ck +(k+2)(k+1)C+2]x* = 0 O This option (CBC₂)+(4C-24₂) +0+1G-4(+2)+1²-0 MWI Mi O This option 60₁x²-0 2C₁-C₂ +6C₂x+2+1)(k-1₂ +k+ 2)(k+1)x-0 2C₂+Co+(1-2k)C +(k+ 2) (k+1)C.]x* = 0
By shifting the summation index of the following added three series Σ n(n-1)₂x + [n n(n-1)C₂x-²-4n1"+ WW2 192 we get: Σ[(k-2)(k-3)Ck +(k+2)(k+1)C+2]x* = 0 O This option (CBC₂)+(4C-24₂) +0+1G-4(+2)+1²-0 MWI Mi O This option 60₁x²-0 2C₁-C₂ +6C₂x+2+1)(k-1₂ +k+ 2)(k+1)x-0 2C₂+Co+(1-2k)C +(k+ 2) (k+1)C.]x* = 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![By shifting the summation index of the following added three series
n(n – 1)C, x" +
n(n – 1)C,x*-3 –) 4nC, 1" +
6C,x"-0
TOO_BL
we get:
E[(k - 2)(k - 3)C
+(k + 2)(k + 1)C+2]x*0
(k + 2)(k + 1)G,.l-0
O This option
This option
2C, + C, +E(1 – 2k)G
+(k + 2) (k + 1)C.al* - 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7f1bda98-f382-4184-97a7-15c19f610c80%2Fbad7dbe1-63d1-4420-a2e4-a8544addb84b%2Fvtifcca_processed.jpeg&w=3840&q=75)
Transcribed Image Text:By shifting the summation index of the following added three series
n(n – 1)C, x" +
n(n – 1)C,x*-3 –) 4nC, 1" +
6C,x"-0
TOO_BL
we get:
E[(k - 2)(k - 3)C
+(k + 2)(k + 1)C+2]x*0
(k + 2)(k + 1)G,.l-0
O This option
This option
2C, + C, +E(1 – 2k)G
+(k + 2) (k + 1)C.al* - 0
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