Mathematical Induction with Sigma notation

[NotChelsea]

Prove by mathematical induction that

n
sigma r^3 = n^2(n+1)^2/4
r = 1

so far I have

1
sigma r^3 = 1^2(1+1)^2/2
r=1

1 = 1(4)/2

1 = 4/2

1 = 2

I'm not sure what to do after this for the k+1 case.

 

[MarkFL]

Hello, and welcome to MHB! (Wave)

Your induction hypothesis $P_n$ is:

$$\sum_{r=1}^{n}\left(r^3\right)=\frac{n^2(n+1)^2}{4}$$

As your induction step, I would suggest adding $$(n+1)^3$$ to both sides...

edit: I just noticed you have an error in your check of the base case $P_1$...can you spot it?

 

[NotChelsea]

ummmm is it because of the "1 = 2"?

 

[Opalg]

The "1=2" happened because you divided by 2 instead of 4.

 

[HOI]

Prove by mathematical induction that

n
sigma r^3 = n^2(n+1)^2/4
r = 1

so far I have

1
sigma r^3 = 1^2(1+1)^2/2
r=1

As Opalg pointed out, in your original formula you have "/4" but have accidently changed that to "/2".

1 = 1(4)/2

1 = 4/2

1 = 2

which is NOT true! But with "4" in the denominator as in the original formula this becomes 1= 1(4)/4 which is true.

I'm not sure what to do after this for the k+1 case.

The reason so many induction problems involve sums is that then the "k+1" case is just the "k" case plus the next term!

With n= k+ 1 you have
[math]\sum_{r=1}^{k+1} r^3= \sum_{r= 1}^k r^3+ (k+1)^3[/math]
[math]= k^2(k+1)^2/4+ (k+1)^3[/math].

Factor [math](k+1)^2[/math] out of that to get
[math]= (k+1)^2(k^2/4+ k+ 1)= (k+1)^2(k^2+ 4k+ 4)/4[/math]

And now you just have to observe that [math]k^2+ 4k+ 4= (k+ 2)^2= ((k+1)+1)^2[/math] so that [math](k+1)^2(k^2+ 4k+ 4)/4= (k+1)^2((k+1)+1)^2/4[/math]
really is "[math]n^2(n+1)^2/4[/math]" with n= k+ 1.

 

[ChelseaL]

ohhhh I see it now. Thank you so much!