- #1
Benny
- 584
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Hello, I am wondering how explicitly a result needs to be shown in induction questions.
For example prove that 1² + 2² +...+n² = (1/6)n(n+1)(2n+1) where n is a natural number. Here is what I did.
1. P(n) is the statement that the sum of the squares of the first n positive integers is (1/6)n(n+1)(2n+1). Show P(2) is true.
Edit: Fixed first sentence of the above paragraph.
2. Assume P(k) is true where k is a positive integer.
[tex]
1^2 + 2^2 + ... + k^2 = \frac{1}{6}k\left( {k + 1} \right)\left( {2k + 1} \right)
[/tex]
3. Show that P(k+1) is true.
[tex]
1^2 + 2^2 + ... + k^2 + \left( {k + 1} \right)^2 = \frac{1}{6}k\left( {k + 1} \right)\left( {2k + 1} \right) + \left( {k + 1} \right)^2
[/tex]
[tex]
= \left( {k + 1} \right)\left[ {\frac{1}{6}k\left( {2k + 1} \right) + \left( {k + 1} \right)} \right]
[/tex]
[tex]
= \frac{1}{6}\left( {k + 1} \right)\left[ {k\left( {2k + 1} \right) + 6\left( {k + 1} \right)} \right]
[/tex]
[tex]
= \frac{1}{6}\left( {k + 1} \right)\left[ {2k^2 + 7k + 6} \right]
[/tex]
[tex]
= \frac{1}{6}\left( {k + 1} \right)\left( {k + 2} \right)\left( {2k + 3} \right)
[/tex]
Is it fine up to the point above? Or do I need to also have the following?
[tex]
= \frac{1}{6}\left( {k + 1} \right)\left( {\left( {k + 1} \right) + 1} \right)\left( {2\left( {k + 1} \right) + 1} \right)
[/tex]
Any help appreciated.
For example prove that 1² + 2² +...+n² = (1/6)n(n+1)(2n+1) where n is a natural number. Here is what I did.
1. P(n) is the statement that the sum of the squares of the first n positive integers is (1/6)n(n+1)(2n+1). Show P(2) is true.
Edit: Fixed first sentence of the above paragraph.
2. Assume P(k) is true where k is a positive integer.
[tex]
1^2 + 2^2 + ... + k^2 = \frac{1}{6}k\left( {k + 1} \right)\left( {2k + 1} \right)
[/tex]
3. Show that P(k+1) is true.
[tex]
1^2 + 2^2 + ... + k^2 + \left( {k + 1} \right)^2 = \frac{1}{6}k\left( {k + 1} \right)\left( {2k + 1} \right) + \left( {k + 1} \right)^2
[/tex]
[tex]
= \left( {k + 1} \right)\left[ {\frac{1}{6}k\left( {2k + 1} \right) + \left( {k + 1} \right)} \right]
[/tex]
[tex]
= \frac{1}{6}\left( {k + 1} \right)\left[ {k\left( {2k + 1} \right) + 6\left( {k + 1} \right)} \right]
[/tex]
[tex]
= \frac{1}{6}\left( {k + 1} \right)\left[ {2k^2 + 7k + 6} \right]
[/tex]
[tex]
= \frac{1}{6}\left( {k + 1} \right)\left( {k + 2} \right)\left( {2k + 3} \right)
[/tex]
Is it fine up to the point above? Or do I need to also have the following?
[tex]
= \frac{1}{6}\left( {k + 1} \right)\left( {\left( {k + 1} \right) + 1} \right)\left( {2\left( {k + 1} \right) + 1} \right)
[/tex]
Any help appreciated.
Last edited: