Mathematical Induction with Sigma notation

In summary, to prove by mathematical induction that \sum_{r=1}^{n} r^3= \frac{n^2(n+1)^2}{4}, you first need to verify the base case P_1, which should result in 1=1. Then, for the induction step, add (n+1)^3 to both sides, factor out (k+1)^2, and observe that k^2+ 4k+ 4= (k+2)^2. This shows that the k+1 case is true, and thus the statement holds for all n.
  • #1
NotChelsea
2
0
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.
 
Last edited:
Physics news on Phys.org
  • #2
Hello, and welcome to MHB! (Wave)

Your induction hypothesis $P_n$ is:

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

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

edit: I just noticed you have an error in your check of the base case $P_1$...can you spot it?
 
  • #3
ummmm is it because of the "1 = 2"?
 
  • #4
The "1=2" happened because you divided by 2 instead of 4.
 
  • #5
NotChelsea said:
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
\(\displaystyle \sum_{r=1}^{k+1} r^3= \sum_{r= 1}^k r^3+ (k+1)^3\)
\(\displaystyle = k^2(k+1)^2/4+ (k+1)^3\).

Factor \(\displaystyle (k+1)^2\) out of that to get
\(\displaystyle = (k+1)^2(k^2/4+ k+ 1)= (k+1)^2(k^2+ 4k+ 4)/4\)

And now you just have to observe that \(\displaystyle k^2+ 4k+ 4= (k+ 2)^2= ((k+1)+1)^2\) so that \(\displaystyle (k+1)^2(k^2+ 4k+ 4)/4= (k+1)^2((k+1)+1)^2/4\)
really is "\(\displaystyle n^2(n+1)^2/4\)" with n= k+ 1.
 
  • #6
ohhhh I see it now. Thank you so much!
 

1. What is mathematical induction with sigma notation?

Mathematical induction with sigma notation is a method of proof used in mathematics to prove statements about a sequence of numbers or an infinite series. It uses the concept of sigma notation, which represents a sum of terms, and the principle of mathematical induction, which states that if a statement is true for a specific number, and it can be proven that it is also true for the next number, then it is true for all subsequent numbers.

2. How is mathematical induction with sigma notation used?

To use mathematical induction with sigma notation, we first prove that a statement is true for the first number in the sequence or series. Then, we assume that the statement is true for the kth number and use this to prove that it is also true for the k+1th number. This step is repeated until we can show that the statement is true for all numbers in the sequence or series.

3. What is the difference between strong and weak induction?

Strong induction is a more generalized form of mathematical induction that allows us to assume that the statement is true for all numbers up to and including the kth number, instead of just the kth number. Weak induction, on the other hand, only allows us to assume that the statement is true for the kth number.

4. Can mathematical induction with sigma notation be used to prove any statement?

No, mathematical induction with sigma notation can only be used to prove statements that follow a specific pattern, such as those involving sequences or series. It cannot be used to prove statements that do not follow a specific pattern.

5. What is the importance of mathematical induction with sigma notation?

Mathematical induction with sigma notation is an important tool in mathematics as it allows us to prove statements about infinite sequences or series, which would otherwise be impossible to prove. It is also widely used in various branches of mathematics, such as number theory, combinatorics, and calculus.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
5
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
901
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
924
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
8
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
875
  • Set Theory, Logic, Probability, Statistics
Replies
13
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
872
Back
Top