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1.2 2,7,18 Prove 13 + 2³ + ... +N³ = [= N(N+1)]³² for all NEN Step 1 show N=1 Step 2 P=1 [HD] [²]²=1 Assume KEN 13³ +2³+ K³ = [ {K/K+1)] ²³ step 3 Apply K+1 15+2³+ +K³ + (K+1)³ = [K(K+1)] ² + (4+1) ³ InductioN (A+)K]" + (+1)(K+1)/K+1) (+)² (K+1)(h+1)K)/(k) + (K+1) / K+1)(K+1) = [ (h+1)] ²³ (1){(k) + (K+1) (K+1) {K+1) I Don't see how to go to ==+/(x+1)X(X+2)] ²

1.2 2,7,18 Prove 13 + 2³ + ... +N³ = [= N(N+1)]³² for all NEN Step 1 show N=1 Step 2 P=1 [HD] [²]²=1 Assume KEN 13³ +2³+ K³ = [ {K/K+1)] ²³ step 3 Apply K+1 15+2³+ +K³ + (K+1)³ = [K(K+1)] ² + (4+1) ³ InductioN (A+)K]" + (+1)(K+1)/K+1) (+)² (K+1)(h+1)K)/(k) + (K+1) / K+1)(K+1) = [ (h+1)] ²³ (1){(k) + (K+1) (K+1) {K+1) I Don't see how to go to ==+/(x+1)X(X+2)] ²

Algebra and Trigonometry (6th Edition)
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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1.2 2,7,18 Prove 13 + 2³ + ... +N³ = [= N(N+1)]³² for all NEN Step 1 show N=1 Step 2 P=1 [HD] [²]²=1 Assume KEN 13³ +2³+ K³ = [ {K/K+1)] ²³ step 3 Apply K+1 15+2³+ +K³ + (K+1)³ = [K(K+1)] ² + (4+1) ³ InductioN (A+)K]" + (+1)(K+1)/K+1) (+)² (K+1)(h+1)K)/(k) + (K+1) / K+1)(K+1) = [ (h+1)] ²³ (1){(k) + (K+1) (K+1) {K+1) I Don't see how to go to ==+/(x+1)X(X+2)] ²
Transcribed Image Text:1.2 2,7,18 Prove 13 + 2³ + ... +N³ = [= N(N+1)]³² for all NEN Step 1 show N=1 Step 2 P=1 [HD] [²]²=1 Assume KEN 13³ +2³+ K³ = [ {K/K+1)] ²³ step 3 Apply K+1 15+2³+ +K³ + (K+1)³ = [K(K+1)] ² + (4+1) ³ InductioN (A+)K]" + (+1)(K+1)/K+1) (+)² (K+1)(h+1)K)/(k) + (K+1) / K+1)(K+1) = [ (h+1)] ²³ (1){(k) + (K+1) (K+1) {K+1) I Don't see how to go to ==+/(x+1)X(X+2)] ²
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