1.2 #23Prove 1³ +2³+ ... +N³ = [ ½/₂2N (N+1)] ² for all NEN71³= 1 [1/2₂ (0 (1+0] ² > ½/2 (2) → ²/2 = 1 ✓show N=12 Assume KENApply +11³ +2³+ ... + K³ = [ ½/₂K/K+D]²213 +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

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 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,...

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