1.2 #2 3 Prove 1³ +2³+...+N³ = [ ½/2N (N+1)] ² for all NEN show N=1 1³= 1 [1/2₂ (1) (1 +D]² >> ½/₂2 (²) → ¾/2=1 ✓ Assume KEN 3 Apply Pht! 3 1 ³+2 ³+ ... + K³ = [ ½/₂K/K+D]² 1³ +2³+ ... + K³ + (K+1) ³ = [/₂ K (K+1)] ² + (+1) = [/2₂(K+1)K] ² + (K+1) [1/₂2(K+1)(h+2)]²
1.2 #2 3 Prove 1³ +2³+...+N³ = [ ½/2N (N+1)] ² for all NEN show N=1 1³= 1 [1/2₂ (1) (1 +D]² >> ½/₂2 (²) → ¾/2=1 ✓ Assume KEN 3 Apply Pht! 3 1 ³+2 ³+ ... + K³ = [ ½/₂K/K+D]² 1³ +2³+ ... + K³ + (K+1) ³ = [/₂ K (K+1)] ² + (+1) = [/2₂(K+1)K] ² + (K+1) [1/₂2(K+1)(h+2)]²
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section: Chapter Questions
Problem 5T
Related questions
Question
![1.2 #2
3
Prove 1³ +2³+ ... +N³ = [ ½/₂2N (N+1)] ² for all NEN
7
1³= 1 [1/2₂ (0 (1+0] ² > ½/2 (2) → ²/2 = 1 ✓
show N=1
2 Assume KEN
Apply +1
1³ +2³+ ... + K³ = [ ½/₂K/K+D]²
2
13 +2²³ + ... + K³ + (k+1) ³ = [ ½/₂K(K+1)] ² + (+1)
= [ ½/₂2(K+1)K] ² + (K+1)
2
[ ¼/2₂ (h+1)(h+z)]²](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe57a7d70-87de-4a1f-8104-5b2578062c6c%2F978c81a5-fe65-4755-8bf3-c49e85f74490%2Fv3yu42_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1.2 #2
3
Prove 1³ +2³+ ... +N³ = [ ½/₂2N (N+1)] ² for all NEN
7
1³= 1 [1/2₂ (0 (1+0] ² > ½/2 (2) → ²/2 = 1 ✓
show N=1
2 Assume KEN
Apply +1
1³ +2³+ ... + K³ = [ ½/₂K/K+D]²
2
13 +2²³ + ... + K³ + (k+1) ³ = [ ½/₂K(K+1)] ² + (+1)
= [ ½/₂2(K+1)K] ² + (K+1)
2
[ ¼/2₂ (h+1)(h+z)]²
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