1. Mathematical induction

Ok, so I'm trying to use mathematical induction to prove 1^3+2^3+...+n^3=[n(n+1)/2]^2 whenever n is a positive integer. I think I have it, but I'm not sure, can any of you help me to make sure I'm on the right track? Thanks!

Basis step: P(1) states that the sum of the first one integer to the third power is [n(n+1)/2]^2. This is true since the sum of the first one integer to the third power is 1. [1(1+1)/2]^2= [1(2)/2]^2=(2/2)^2=1

Induction Step: Assume P(n) is true for a positive integer n that is

[2(2+1)/2]^2 [3(3+1)/2]^2
[2(3)/2]^2 [3(4)/2]^2
(6/2)^2 (12/2)^2
3^2 6^2
9=1^3+2^3 36=1^3+2^3+3^3
9=1+8 =9 36=1+8+27=36

Conclusion: The sum of the first n positive integers to the third power is given by [n(n+1)/2]^2 for all integers n>=1

2. Originally Posted by sjenkins
Ok, so I'm trying to use mathematical induction to prove 1^3+2^3+...+n^3=[n(n+1)/2]^2 whenever n is a positive integer. I think I have it, but I'm not sure, can any of you help me to make sure I'm on the right track? Thanks!

Basis step: P(1) states that the sum of the first one integer to the third power is [n(n+1)/2]^2. This is true since the sum of the first one integer to the third power is 1. [1(1+1)/2]^2= [1(2)/2]^2=(2/2)^2=1

Induction Step: Assume P(n) is true for a positive integer n that is

[2(2+1)/2]^2 [3(3+1)/2]^2
[2(3)/2]^2 [3(4)/2]^2
(6/2)^2 (12/2)^2
3^2 6^2
9=1^3+2^3 36=1^3+2^3+3^3
9=1+8 =9 36=1+8+27=36

Conclusion: The sum of the first n positive integers to the third power is given by [n(n+1)/2]^2 for all integers n>=1
um, no. you are not doing mathematical induction here

if you assume P(n) is true (by the way, you need to state what P(n) is) then you assume:

$\displaystyle 1^3 + 2^3 + \cdots + n^3 = \Bigg[ \frac {n(n + 1)}2 \Bigg]^2$

now you need to show P(n + 1) holds. that is, you need to show that P(n) holds implies $\displaystyle 1^3 + 2^3 + \cdots (n + 1)^3 = \Bigg[ \frac {(n + 1)(n + 2)}2\Bigg]^2$

Hint: add $\displaystyle (n + 1)^3$ to both sides of P(n) then expand and simplify. or alternatively, you may find it easier to work the other way, start by expanding the right side of the P(n + 1) statement, and show that you can get the left side

3. Is there any way someone can explain this further to me? I am obviously not getting it.

4. Originally Posted by sjenkins
Is there any way someone can explain this further to me? I am obviously not getting it.
since we have verified the base case, assume P(n) is true. now we need to show P(n + 1) is true.

since P(n) is true:

$\displaystyle 1 + 2^3 + \cdots + n^3 = \Bigg[ \frac {n(n + 1)}2 \Bigg]^2$

add $\displaystyle (n + 1)^3$ to both sides

$\displaystyle \Rightarrow 1 + 2^3 + \cdots + n^3 + (n + 1)^3 = \Bigg[ \frac {n(n + 1)}2 \Bigg]^2 + (n + 1)^3$

now, simplify the right hand side to get $\displaystyle \Bigg[ \frac {(n + 1)(n + 2)}2 \Bigg]^2$

that would prove: $\displaystyle 1 + 2^3 + \cdots + n^3 + (n + 1)^3 = \Bigg[ \frac {(n + 1)(n + 2)}2 \Bigg]^2$

which is exactly the statement P(n + 1)

5. Thank you so much for the help, now I can see!

6. Originally Posted by sjenkins
Thank you so much for the help, now I can see!
good