Mathematical Induction Midwestern State University

Why the droplet Template??? Because Mathematical induction makes most students cry! Will try to make it painless.

Principle of Mathematical induction Let S(n) be a statement involving an integer n. Suppose that, for some fixed integer n0, S(n0) is true (i.e. S(n) is true if n = n0 ) When integer k >= n0 & S(k) is true, then S(k+1) is true The S(n) is true for all integers n >= n0

Example S(n): 1 + 2 + 3 +… + n = (n (n + 1)) / 2 n=4  (4*5)/2 = 10

Induction Says…. S(n): 1 + 2 + 3 +… + n = (n (n + 1)) / 2 Show that S(n) is true for some (small) integer k That means it is true at least sometimes! Since S(n) is true for some integer k, show it is also true for k+1 That means it is true for all integers >= k

Lets Prove it using induction S(n): 1 + 2 + 3 +… + n = (n (n + 1)) / 2 Lets first show S(n) is true for some small integer. In this case lets use 1. If we plug in 1 we get 1 = (1(1+1))/2 and hence we see that 1 =1 This means that 1 works in the formula for n

Continuing S(n): 1 + 2 + 3 +… + n = (n (n + 1)) / 2 Now assume that n works. So giving that assumption lets see if S(n+1) works. We can do this by plugging in n+1 for n in the formula. SO! We start with the LHS and generate the RHS using pure algebra! Capice!? S(n+1): 1 + 2 + 3 +… + n + n+1= (n (n+1))/2+ n+1 = (n (n+1))/2+ 2( n+1)/2 = ((n (n+1))+ 2( n+1))/2 = ((n + 2)( n+1))/2 = ((n + 1)+ 1)( n+1))/2 = S (n+1) QED

To REPEAT MYSELF Show that n = 1 works in the formula Assume that n works in the formula and show with this assumption that n+1 works This will prove that ALL of the n values will work

another Problem Prove 1 + 3 + 5 + . . . + (2n-1) =n2 Clearly 1 works so assume that k works so lets prove that k+1 will work 1 + 3 + 5 +…+ (2k-1) + 2(k+1)-1= k2 + 2k + 1 = (k+1)2 QED

IN Class Exercises Prove: For n > 1, 1+4+9+…+n2= n(n+1)(2n+1)/6

Homework Page 84 – Problems 12 & 13