** Claim: **

* Proof. * For , we have 1 on the left, and on the right . Thus, the formula is true for .

Assume then that it is true for some . Then,

Thus, if the formula is true for then it is true for . Since we established it is true for , we have that it is true for all

I don’t understand the 3rd step, please somebody explain me. Thanks.

The numbers in the series alternate between negative and positive. So if you pick one number from the series and it’s negative, the next number will be positive.

Hi Luis. He used the identity: (1+2+…+k) = k(k+1)/2 (this appears in part (a) of the exercise); and he put the factor (-1)^(k+2) out. Notice that you have [(-1)^(k+1)] * [k(k+1)/2], so when you put (-1)^(k+2) out, what stays inside the parenthesis is: [(-1)^(-1)] * [k(k+1)/2]. But (-1)^(-1) = -1. Hence you get, – k(k+1)/2.

Remark: remember that (a^b) * (a^c) = a^(b+c).

Hello. In the 3rd step: How did you go from (-1)^(k+1) to (-1)^(k+2)?

We have [(-1)^(k+1)] * [k(k+1)/2], but i do not see from where i can get (-1)^(k+2) and put out, and therefore have [(-1)^(-1)] * [k(k+1)/2]. Could you please explain?