Section 3.2

• 4. Basis(n=0): 2 = (1-(-7))/4 = 2.

  Ind(n>0): 
  n+1	  i    n       i       n+1
  SUM 2 (-7)  = SUM 2(-7)	+ 2(-7)	  
  i=0	      i=0
  
  		      n+1	 n+1
              = 1 - (-7)	  + 2(-7)   
         	      --------
  		 4    
  
  		      n+2
              = 1 - (-7) 	 
    	      ---------  
                   4 
  
  

• 6. (n=1) 1/2, (n=2) 2/3, (n=3) 3/4. Looks like

   n     1        n
  SUM   ---    = ---
  i=1   i(i+1)   n+1  
  
  Basis(n=1) 1/2 = 1/2
  
  Ind(n>1)  
  
  n+1   1        n    1	       1     	 n 	  1    	   (n+1)
  SUM -----  =  SUM -----  + ---------- = --- + ---------  = -----
  i=1 i(i+1)    i=1 i(i+1)   (n+1)(n+2)	n+1   (n+1)(n+2)   (n+2)
  

• 8.
• 14. Basis(n=2): 2! = 2 < 2^2 = 4. Ind(n>2): (n+1)! = (n+1)n! < (n+1)n^n < (n+1)(n+1)^n = (n+1)^(n+1).
• 18.
• 20. Basis(n=0): 3 divides 0. Ind(n>0): (n+1)^3 + 2(n+1) = n^3 + 2n + 3(n^2 + n + 1). 3 divides n^3+2n by inductive hypothesis, so 3 divides whole thing.
• 26. Let S be set with n elements, and T = S U {a}, where a is not in S. Number of 3-element subsets of T is equal to the number of such subsets in S, plus the number of 2-element subsets in S, because we just union a with these. By Exercise 25, there are n(n-1)/2 subsets of 2 elements in S. So the number we are after, using the inductive hypothesis, is n(n-1)(n-2)/6 + n(n-1)/2 = (n+1)n(n-1)/6.
• 34. A^(n+1) = A A^n = [a 0; 0 b][a^n 0; 0 b^n] = [a^n 0; 0 b^n].
• 40.
• 44. Basis was for n=0. Inductive step assumed n >= 1. Doesn't include 0!