induction

Question 1

sum of all positive integers including n = (n * (n +1)) /2

Answer 1

induction hypothesis (n* (n + 1)) /2
1) base s(1) = (1 * (1 + 1)) /2
                   =  (1 * 2) /2
                   =   2/2
                   = 1
2) induction case
    s (k+1) = ((1 * (1 + 1)) /2) + (k+1) 
common denomator
               = ((k(k + 1)) /2) + ((2(k+1))/2)
               = ((k(k +1) + (2(k+1)))/2
factor out (k+1)
               = ((k+1)(k+2))/2 
turn k+2 to k+1+1
                = ((k+1)((k+1)+2))/2
                                                              []

Question 2

(*) For n > 1, 2 + 2^2 + 2^3 + 2^4 + … + 2^n = 2^n+1 – 2

Answer 2

base case (1)
s(1) = 2^1 = 2 
     = 2^(n+1) – 2 = 2^(1+1) – 2 = 2^2 – 2 = 4 – 2 = 2
2) induction case
Let n = k + 1.

2 + 2^2 + 2^3 + 2^4 + ... + 2^k + 2^(k+1) 
     = [2 + 2^2 + 2^3 + 2^4 + ... + 2^k] + 2^(k+1) 
     = [2^(k+1) – 2] + 2^(k+1) 
there are 2 * 2^(k+1) 
     = 2 × 2^(k+1) – 2  
raise 2^1 is 2
     = 2^1 × 2(k+1) – 2 
multiply powers
     = 2^(k+1+1) – 2  
transitive plus brackets 
     = 2^(k+1)+1 – 2

Question 3

For n > 1, 1×2 + 2×3 + 3×4 + … + (n)(n+1) = (n)(n+1)(n+2)/3

Answer 3

Let n = 1
Then the left-hand side of () is 1×2 = 2
and the right-hand side of () is (1)(2)(3)/3 = 2

Let n = k + 1. The left-hand side of (*) then gives us:

1×2 + 2×3 + 3×4 + … + (k)(k+1) + (k+1)((k+1)+1)
= [1×2 + 2×3 + 3×4 + … + (k)(k+1)] + (k+1)((k+1)+1)
= [(k)(k+1)(k+2)/3] + (k+1)((k+1)+1)

     = (k)(k+1)(k+2)/3 + (k+1)(k+2) 
     = (k)(k+1)(k+2)/3 + 3(k+1)(k+2)/3 
     = (k+3)(k+1)(k+2)/3 
     = (k+1)(k+2)(k+3)/3 
     = (k+1)((k+1)+1)((k+1)+2)/3