Induction Method for Proving 1² + 2² +...+n² = (1/6)n(n+1)(2n+1)

[Benny]

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.

$$1^2 + 2^2 + ... + k^2 = \frac{1}{6}k\left( {k + 1} \right)\left( {2k + 1} \right)$$


3. Show that P(k+1) is true.

$$1^2 + 2^2 + ... + k^2 + \left( {k + 1} \right)^2 = \frac{1}{6}k\left( {k + 1} \right)\left( {2k + 1} \right) + \left( {k + 1} \right)^2$$

$$= \left( {k + 1} \right)\left[ {\frac{1}{6}k\left( {2k + 1} \right) + \left( {k + 1} \right)} \right]$$

$$= \frac{1}{6}\left( {k + 1} \right)\left[ {k\left( {2k + 1} \right) + 6\left( {k + 1} \right)} \right]$$

$$= \frac{1}{6}\left( {k + 1} \right)\left[ {2k^2 + 7k + 6} \right]$$

$$= \frac{1}{6}\left( {k + 1} \right)\left( {k + 2} \right)\left( {2k + 3} \right)$$

Is it fine up to the point above? Or do I need to also have the following?

$$= \frac{1}{6}\left( {k + 1} \right)\left( {\left( {k + 1} \right) + 1} \right)\left( {2\left( {k + 1} \right) + 1} \right)$$

Any help appreciated.

 

[vincentchan]

your last step is absolutely nessisory (sry, don't know how to spell this word)... The reason is: you must show in the P(n+1) case, your formulas is the exactly the same as P(n) with n replaced by (n+1)...

 

[Spectre5]

first you must show it is true when n = 1

which is easy enough, 1 = (1/6)*1*2*3

then do n+1 as you did

I personally would write that last line you wrote to make it very clear that n was replaced with K+1

 

[Benny]

Cool, thanks for the help vincent and spectre.