Proving the Formula Using Mathematical Induction

In summary, the formula for summing the fifth powers of positive integers from 1 to n can be proven using mathematical induction, with the base case being n=1 and the assumption that it holds for k. By manipulating the equation and adding (k+1)^5 to both sides, the equation can be simplified to \frac{(k+1)^2(k+2)^2(2(k+1)^2+2k+1)}{12}=\frac{k^2 (2k^2+2k-1)+12(k+1)^3}{12}. From here, the two sides can be expanded and the common factor (k+1)^2 can be factored out, showing that the two sides
  • #1
themadhatter1
140
0

Homework Statement


Use mathematical induction to prove the formula for every positive integer n.

[tex]\sum_{i=1}^{5}i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}[/tex]

Homework Equations




The Attempt at a Solution



I know this will be true because the RHS is just your standard sums of powers of integers to the fifth formula, but nevertheless I have to prove it...

I know n=1 will work so I set n=k

[tex]\sum_{k=1}^{5}k^5=\frac{k^2(k+1)^2(2k^2+2k-1)}{12}[/tex]

I take the RHS and set k=(k+1)


[tex]=\frac{(k+1)^2(k+2)^2(2(k+1)^2+2(k+1)-1)}{12}[/tex]
[tex]=\frac{(k+1)^2(k+2)^2(2(k+1)^2+2k+1)}{12}[/tex]

If the test works for all the terms then it must be correct for the first 5 terms so I can eliminate the summation and just have the [tex]k^5[/tex]

so then I'd have

[tex](k+1)^5+\frac{k^2(k+1)^2(2k^2+2k-1)}{12}[/tex]
[tex]=\frac{k^2(k+1)^2(2k^2+2k-1)+12(k+1)^5}{12}[/tex]

From here I'm not sure how to get that to equal this:

[tex]\frac{(k+1)^2(k+2)^2(2(k+1)^2+2k+1)}{12}[/tex]
 
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  • #2
themadhatter1 said:

Homework Statement


Use mathematical induction to prove the formula for every positive integer n.

[tex]\sum_{i=1}^{5}i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}[/tex]

Homework Equations

The Attempt at a Solution



I know this will be true because the RHS is just your standard sums of powers of integers to the fifth formula, but nevertheless I have to prove it...

I know n=1 will work so I set n=k

[tex]\sum_{k=1}^{5}k^5=\frac{k^2(k+1)^2(2k^2+2k-1)}{12}[/tex]

I take the RHS and set k=(k+1)[tex]=\frac{(k+1)^2(k+2)^2(2(k+1)^2+2(k+1)-1)}{12}[/tex]
[tex]=\frac{(k+1)^2(k+2)^2(2(k+1)^2+2k+1)}{12}[/tex]

If the test works for all the terms then it must be correct for the first 5 terms so I can eliminate the summation and just have the [tex]k^5[/tex]

so then I'd have

[tex](k+1)^5+\frac{k^2(k+1)^2(2k^2+2k-1)}{12}[/tex]
[tex]=\frac{k^2(k+1)^2(2k^2+2k-1)+12(k+1)^5}{12}[/tex]

From here I'm not sure how to get that to equal this:

[tex]\frac{(k+1)^2(k+2)^2(2(k+1)^2+2k+1)}{12}[/tex]

Your first formula as written is incorrect. The sum from 1 to 5 is not different for every positive integer, n. Your upper index should be n and not 5.

To use induction you must:
1. Verify the base case.
2. Assume that your case holds for some number, k.
3. Show that it holds for k+1. (IF it is true for k, THEN it is true for k+1)

Try adding [tex](k+1)^5[/tex] to both sides of the equation and manipulate the right side until you get the proper form.
 
  • #3
malicx said:
Your first formula as written is incorrect. The sum from 1 to 5 is not different for every positive integer, n. Your upper index should be n and not 5.

To use induction you must:
1. Verify the base case.
2. Assume that your case holds for some number, k.
3. Show that it holds for k+1. (IF it is true for k, THEN it is true for k+1)

Try adding [tex](k+1)^5[/tex] to both sides of the equation and manipulate the right side until you get the proper form.

My first formula is written exactly as printed from my textbook so I know it is correct. I know how to solve with induction my question is only one of algebra.

I need to get

[tex]\frac{(k+1)^2(k+2)^2(2(k+1)^2+2k+1)}{12}=\frac{k^2(k+1)^2(2k^2+2k-1)+12(k+1)^5}{12}[/tex]

The one on the left is where I subbed in k=(k+1) the one on the right is the equation I got from adding (k+1)5 to. I know they are equal because you can plug in a value and get the same number on both sides. How can you get these equations into the same form.
 
  • #4
Failing anything clever, you could always expand and collect both numerators to see they are both:

2k6+18k5+65k4+120k3+119k2+60k+12
 
  • #5
themadhatter1 said:
[tex]\frac{(k+1)^2(k+2)^2(2(k+1)^2+2k+1)}{12}=\frac{k^2(k+1)^2(2k^2+2k-1)+12(k+1)^5}{12}[/tex]
That is correct. Note that both sides have the common factor (k+1)^2, so you have to prove the simpler equality

[tex]\frac{(k+2)^2(2(k+1)^2+2k+1)}{12}=\frac{k^2 (2k^2+2k-1)+12(k+1)^3}{12}[/tex].

After that you can follow LCKurtz suggestion to expand both sides and see if they are identical.

ehild
 
  • #6
Alright, that works.

Thanks.
 
  • #7
themadhatter1 said:
My first formula is written exactly as printed from my textbook so I know it is correct.
At the risk of sounding like a stickler/jerk, the formula is NOT correct, even if it is what is listed in the book. Typically, a formula in n will be given when summing from 1 to n, but that is summing from 1 to 5.
 
  • #8
You are right, it must be a typo. So the correct form is

[tex]
\sum_{i=1}^{n}i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}
[/tex]

ehild
 

What is mathematical induction?

Mathematical induction is a method of mathematical proof used to establish the truth of a mathematical statement for all values of a variable. It involves proving that the statement holds for the first value of the variable, and then showing that if it holds for any given value, it must also hold for the next value.

Why is mathematical induction important?

Mathematical induction is important because it is one of the fundamental tools used in mathematics to prove theorems and establish the validity of mathematical statements. It is also used in many areas of computer science and engineering, making it an essential concept to understand for problem-solving and critical thinking.

How does mathematical induction differ from other methods of proof?

Unlike other methods of proof, such as direct proof or proof by contradiction, mathematical induction does not require a specific value to be proven. Instead, it proves that a statement is true for all values of a variable, making it a powerful and versatile tool for tackling a wide range of mathematical problems.

What are the steps involved in using mathematical induction?

The general steps for using mathematical induction are as follows:

  1. Prove that the statement is true for the first value of the variable.
  2. Assume that the statement is true for any arbitrary value of the variable.
  3. Use this assumption to prove that the statement is also true for the next value of the variable.
  4. Conclude that the statement must be true for all values of the variable by the principle of mathematical induction.

What are some common mistakes to avoid when using mathematical induction?

Some common mistakes to avoid when using mathematical induction include:

  • Not proving the base case (first value) of the statement.
  • Assuming that the statement is true for a specific value of the variable, rather than an arbitrary value.
  • Using circular reasoning or making assumptions that are not justified.
  • Not being clear and precise in the language and notation used in the proof.
  • Forgetting to include all necessary steps or not clearly explaining each step in the proof.

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