help with an example (induction)

[Clifford]

Prove that 1+2+..+n = n(n+1)/2

Let P(n) be the statement: 1+2+..+n = n(n+1)/2

the first step is to solve for P(1), which both sides will be 1 if you work through the math.

the next step is to assume that P(k) is true and then lastly we need to prove it true for P(k+1).

The math in the example shows:

1+2+..+k+(k+1) = k(k+1)/2 + (k+1)
1+2+..+k+(k+1) = (k+1)(k+2)/2

I don't understand how they went from the first step to the second step. Where does the (k+2) come in, and I don't understand how its true.

 

[Deleted member 4993]

Prove that 1+2+..+n = n(n+1)/2

Let P(n) be the statement: 1+2+..+n = n(n+1)/2

the first step is to solve for P(1), which both sides will be 1 if you work through the math.

the next step is to assume that P(k) is true and then lastly we need to prove it true for P(k+1).

The math in the example shows:

1+2+..+k+(k+1) = k(k+1)/2 + (k+1) = (k+1) {k/2 + 1} = (k+1){(k+2)/2}
1+2+..+k+(k+1) = (k+1)(k+2)/2

I don't understand how they went from the first step to the second step. Where does the (k+2) come in, and I don't understand how its true.

 

[Denis]

If Mr K's is too hard for you to follow, you can do it the old-fashioned way:

k(k + 1) / 2 + (k + 1)

= [k(k + 1) + 2(k+1)] / 2

= (k^2 + k + 2k + 2) / 2

= (k^2 + 3k + 2) / 2 ; now factor that:

= (k + 1)(k + 2) / 2 : kapish?

 

[Clifford]

Thanks, I understand now how you get the (k+1)(K+2) / 2 now, however I do not understand how this shows that P(k+1) is true. Could somebody explain how this result shows/proves that P(k+1) is true?

 

[stapel]

I do not understand how this shows that P(k+1) is true.

Hint: What would "P(k + 1)" look like? (Plug "k + 1" in for "n", and look at the result.) :wink:

Eliz.