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Το show 1² + 2² + 3² +4² +5² + To proove by induction; + n² n(n+1)(2n+1) 6

Το show 1² + 2² + 3² +4² +5² + To proove by induction; + n² n(n+1)(2n+1) 6

College Algebra
College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter9: Sequences, Probability And Counting Theory
Section: Chapter Questions
Problem 45RE: Use the Binomial Theorem to write the first three terms of (2a+b)17.
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1² + 2² + 3² + MEK By Induction, 2 + K² + (k+ 11² EIN 1² +2²+ = + n² = = = so statement is true = (K+1) { 2K²+ 7k+6} 6 - (K+1) { K(2K+1) + 6(k+1}} 6 = (K+1) { 2K²+k+ 6K+6} 6 = · (K+¹) { 2K² +3K+ 4K+6} G (k+1) { (2k + 3) (k+2)} 6 (k+1) (K+1+1) (2(K+1)+1) 6 for n = (k+1) n(n+1) (2n+1) 6 when Hence prooved. true for
Transcribed Image Text:1² + 2² + 3² + MEK By Induction, 2 + K² + (k+ 11² EIN 1² +2²+ = + n² = = = so statement is true = (K+1) { 2K²+ 7k+6} 6 - (K+1) { K(2K+1) + 6(k+1}} 6 = (K+1) { 2K²+k+ 6K+6} 6 = · (K+¹) { 2K² +3K+ 4K+6} G (k+1) { (2k + 3) (k+2)} 6 (k+1) (K+1+1) (2(K+1)+1) 6 for n = (k+1) n(n+1) (2n+1) 6 when Hence prooved. true for
To show To proove 1² + 2² + 3² +4² +5². by Base Case : 1 Let >> 2 1²+ 2² + 3² + n = 1 So statement inductive case : Now Now n(n+1)(2n+1) 6 Let statement we will n=k+1 induction; when true for statement is true for 1² + 2² + 3² + show is true true 1²+ 2² + 3² + n². = = = that = far for n(n+1)(2+1) 6 1² = n = 1 1(1+1)(2+1) 6 + n² 2 +k² MEK n = K state ment n≤k 2 + k² + (k+1) ² k(k+1)(2k+1) 6 (x+1) { = K(K+1)(2K+1) 6 is true for + (k+1)² K(2K+1) + (k+1) } By 0 all
Transcribed Image Text:To show To proove 1² + 2² + 3² +4² +5². by Base Case : 1 Let >> 2 1²+ 2² + 3² + n = 1 So statement inductive case : Now Now n(n+1)(2n+1) 6 Let statement we will n=k+1 induction; when true for statement is true for 1² + 2² + 3² + show is true true 1²+ 2² + 3² + n². = = = that = far for n(n+1)(2+1) 6 1² = n = 1 1(1+1)(2+1) 6 + n² 2 +k² MEK n = K state ment n≤k 2 + k² + (k+1) ² k(k+1)(2k+1) 6 (x+1) { = K(K+1)(2K+1) 6 is true for + (k+1)² K(2K+1) + (k+1) } By 0 all
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