1.2 #2 2 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 2 Assume KEN Apply Yht! 3 1³ +2³² + ... + K³ = [ ½/2₂K (K+D]² 13³ +2³3 + ... + K³ + (K+1) ³ = [ ½/₂K(K+1)] ² + (K+1) = [ ½/₂(K+1)K]²+ (K+1) 2 [ ½/2[h+1)(h+z)] ²
1.2 #2 2 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 2 Assume KEN Apply Yht! 3 1³ +2³² + ... + K³ = [ ½/2₂K (K+D]² 13³ +2³3 + ... + K³ + (K+1) ³ = [ ½/₂K(K+1)] ² + (K+1) = [ ½/₂(K+1)K]²+ (K+1) 2 [ ½/2[h+1)(h+z)] ²
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
![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%2Fa2cdb6af-9b3b-475f-a155-8d2b254b2c01%2F6exotud_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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