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Mathematical Induction
Midwestern State University
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Why the droplet Template???
Because Mathematical induction makes most students cry! Will try to make it painless.
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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
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Example S(n): 1 + 2 + 3 +… + n = (n (n + 1)) / 2 n=4 (4*5)/2 = 10
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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
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Lets Prove it using induction
S(n): … + 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
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Continuing S(n): … + 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): … + 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
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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
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another Problem Prove (2n-1) =n2 Clearly 1 works so assume that k works so lets prove that k+1 will work …+ (2k-1) + 2(k+1)-1= k2 + 2k + 1 = (k+1)2 QED
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IN Class Exercises Prove: For n > 1, 1+4+9+…+n2= n(n+1)(2n+1)/6
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Homework Page 84 – Problems 12 & 13
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