Question

Peraturan :
1. Dilarang keras menjawab asal-asalan / spam !
2. Menggunakan cara
3. Jika tidak paham silah kan bertanya di kolom komentar yang sudah disediakan jangan di kolom jawaban !

Soal
1³ + 2³ + 3³ + ... + n³ = (1 + 2 + 3 + ... + n)³

Answer 1

S(n): 1³ + 2³ + 3³ +...+ n³ = (1+ 2+ 3+...+ n)² = [n.(n+1)/2]² 

a) S(n) ...n = 1: 

S(1): 1³ = 1² => S(1) 

b) S(n) --> S(n+1) 

1³ + 2³ + 3³ +...+ n³ = (1+ 2+ 3+...+ n)² 

1+ 2+ 3+...+ n = n(n+1)/2 


1³ + 2³ + 3³ +...+ n³ = (n(n+1)/2)². 


1³ + 2³ +...+ n³+ (n+1)³ = (n(n+1)/2)² + (n+1)³ 
= (n+1)²(n/2)² + (n+1)²(n+1) 
= (n+1)²((n/2)²+ (n+1)) 
= (n+1)²(n²/2² + n +1 ) 
= (1/2²)(n+1)²(n²+ 2²n + 2²) 
= (1/2²)(n+1)²(n²+ 2*2n + 2²) 
= (1/2²)(n+1)²(n+ 2)² 
= ((n+1)(n+ 2)/2)² 
= (1+ 2+ 3 +...+ n + (n+1))². 



1³ + 2³ +...+ n³+ (n+1)³ = (1+ 2+ 3 +...+ n + (n+1))². 

..........................................................................................

induksi ?

(1+2+3+...+n)² = [n(n+1)/2]² 

for n = 1 
= 1³ = 1 
= [ 1² (1 + 1)² ] / 4 = 4/4 = 1 
 
then 
1³ + 2³ + 3³ + ... + k³ = [ k² (k + 1)² ] /4 

n = k + 1 
= 1³ + 2³ + 3³ + ... + k³ + (k + 1)³ 
=[ k² (k + 1)² ] /4 + (k + 1)³ 
= [ k² (k + 1)² + 4(k + 1)³ ] /4 
= [ k² (k + 1)² + 4(k + 1)³ ] /4 
= [ k⁴ + 2k³ + k² + 4k³ + 12k² + 12k + 4 ] /4 
= [ k⁴ + 6k³ + 13k² + 12k + 4 ] /4 

 
= [ (k + 1)² (k + 2)² ] /4 
= [ (k² + 2k + 1)(k² + 4k + 4) ] /4 
= [ k⁴ + 6k³ + 13k² + 12k + 4 ] /4 


P(k) = P(k + 1) 

so by mathematical induction, the statement is true