Homework for week 1. 

P 15 # 1, 4fg, 9fg, 12, 16ef, 20ab, 28ef, 29ef, 33b.

p 26 # 3a, 8a, 16, 29, 30, 33. 

Also, a mathematical induction problem: Prove that if n is a positive integer, then 

1*2+2*3+etc.+(n-1)*n=(n-1)*n*(n+1)/3. 

Thus, the case n=1 of this statement says that 0=0, 

while the case n=3 of the statement says that 1*2+2*3=2*3*4/3. 

For this first induction proof, you get a lot of guidance. 

Do it just like the following worked example: 

Prove that if n is a positive integer, then 1+2+...+n=(n*(n+1)/2. 

Proof. First note that the case n=1 of the statement to be proved says that 1=1*2/2. 

Although it is not essential to the proof, note also that the case n=2 says that 
1+2=2*3/2, and the case n=3 says that 1+2+3=3*4/2, 

and these next two statements are also true. 

Now assume that the case n=k of the statement is true, that is, assume that 

1+2+...+k=k*(k+1)/2. Then we claim the next case, n=k+1, is also true, 

that is, we claim that (1+2+...+k)+(k+1)=(k+1)*(k+2)/2. 

To justify this claim, we calculate: 

(1+2+...+k)+(k+1)=k*(k+1)/2+(k+1) (by our assumption), 
                 =k*(k+1)/2+(k+1)*2/2 (algebra)
                 =(1/2)*(k*(k+1)+2*(k+1)) (more algebra)
                 =(1/2)*(k+2)*(k+1) (distributive rule)
                 =(k+1)*(k+2)/2 as required. 

This completes the proof by mathematical induction that for all positive integers n, 

1+2+...+n=n*(n+1)/2. 

(Your actual proof should use the same words, phrasing etc. 
(there will be room for creativity aplenty, later), 
but the calculation will have to be different, because the problem itself is different. 
Your calculation will start with the line 

(1*2+...+(n-1)*n)+n*(n+1)=(n-1)*n*(n+1)/3+n*(n+1), 

and will end with the line 

                         =n*(n+1)*(n+2)/3 as required.