Finding Formula for partial sums of series.

[teroenza]

Homework Statement

I have the series 1^3+2^3+3^3...n^3, and need to find a formula containing n to represent the sum of the nth terms. The motivation is to find a conjecture, which I can then prove using mathematical induction.

The Attempt at a Solution

I see that

n=1 , 1^3=1

n=2 , 1^3+2^3=9

n=3, 1^3+2^3+3^3=36

n=4, sum=100
n=5 sum = 225. But I cannot come up with a formula to represent the sums. I think this is calculus II series material, but I'm not sure where to start. Could someone point me in the right direction? I used a computer to arrive at

1^3+2^3+3^3...n^3=(1/4)*n^2*(n+1)^2

but do not understand how to get there myself.

 

[fauboca]

Homework Statement

I have the series 1^3+2^3+3^3...n^3, and need to find a formula containing n to represent the sum of the nth terms. The motivation is to find a conjecture, which I can then prove using mathematical induction.

The Attempt at a Solution

I see that

n=1 , 1^3=1

n=2 , 1^3+2^3=9

n=3, 1^3+2^3+3^3=36

n=4, sum=100
n=5 sum = 225. But I cannot come up with a formula to represent the sums. I think this is calculus II series material, but I'm not sure where to start. Could someone point me in the right direction? I used a computer to arrive at

1^3+2^3+3^3...n^3=(1/4)*n^2*(n+1)^2

but do not understand how to get there myself.

1=1^2

9 = x^2

36 = y^2

100 = z^2

etc what are x,y,z? notice something?

 

[teroenza]

Ok.

n= 1 2 3 4 5
sum= 1 3^2 6^2 10^2 15^2

(sum)^1/2= (n/2)*(n+1)

So [(n/2)*(n+1)]^2.