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mathematical induction

Mathematical induction has the following steps: 1) Prove that the statement is true for the base case (usually n=1). 2) Assume the statement is true for some integer k. 3) Using the assumption from step 2, prove the statement is true for k+1. 4) By proving the statement true for n=1 and showing that if it is true for k then it is true for k+1, the statement is true for all positive integers n.

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Mathematical Induction
Mathematical Induction Step 1: Prove the result is true for n = 1 (or whatever the first term is)
Mathematical Induction Step 1: Prove the result is true for n = 1 (or whatever the first term is) Step 2: Assume the result is true for n = k, where k is a positive integer (or another condition that matches the question)
Mathematical Induction Step 1: Prove the result is true for n = 1 (or whatever the first term is) Step 2: Assume the result is true for n = k, where k is a positive integer (or another condition that matches the question) Step 3: Prove the result is true for n = k + 1
Mathematical Induction Step 1: Prove the result is true for n = 1 (or whatever the first term is) Step 2: Assume the result is true for n = k, where k is a positive integer (or another condition that matches the question) Step 3: Prove the result is true for n = k + 1 NOTE: It is important to note in your conclusion that the result is true for n = k + 1 if it is true for n = k
Mathematical Induction Step 1: Prove the result is true for n = 1 (or whatever the first term is) Step 2: Assume the result is true for n = k, where k is a positive integer (or another condition that matches the question) Step 3: Prove the result is true for n = k + 1 NOTE: It is important to note in your conclusion that the result is true for n = k + 1 if it is true for n = k Step 4: Since the result is true for n = 1, then the result is true for all positive integral values of n by induction
1 e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3
1 e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3 Step 1: Prove the result is true for n = 1
1 e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3 Step 1: Prove the result is true for n = 1 LHS  12 1
1 e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3 Step 1: Prove the result is true for n = 1 1 LHS  12 RHS  12  12  1 3 1 1  113 3 1
1 e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3 Step 1: Prove the result is true for n = 1 1 LHS  12 RHS  12  12  1 3 1 1  113 3 1  LHS  RHS
1 e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3 Step 1: Prove the result is true for n = 1 1 LHS  12 RHS  12  12  1 3 1 1  113 3 1  LHS  RHS Hence the result is true for n = 1
1 e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3 Step 1: Prove the result is true for n = 1 1 LHS  12 RHS  12  12  1 3 1 1  113 3 1  LHS  RHS Hence the result is true for n = 1 Step 2: Assume the result is true for n = k, where k is a positive integer
1 e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3 Step 1: Prove the result is true for n = 1 1 LHS  12 RHS  12  12  1 3 1 1  113 3 1  LHS  RHS Hence the result is true for n = 1 Step 2: Assume the result is true for n = k, where k is a positive integer 1 i.e. 12  32  52    2k  1  k 2k  12k  1 2 3
1 e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3 Step 1: Prove the result is true for n = 1 1 LHS  12 RHS  12  12  1 3 1 1  113 3 1  LHS  RHS Hence the result is true for n = 1 Step 2: Assume the result is true for n = k, where k is a positive integer 1 i.e. 12  32  52    2k  1  k 2k  12k  1 2 3 Step 3: Prove the result is true for n = k + 1
1 e.g.i  1  3  5    2n  1  n2n  12n  1 2 2 2 2 3 Step 1: Prove the result is true for n = 1 1 LHS  12 RHS  12  12  1 3 1 1  113 3 1  LHS  RHS Hence the result is true for n = 1 Step 2: Assume the result is true for n = k, where k is a positive integer 1 i.e. 12  32  52    2k  1  k 2k  12k  1 2 3 Step 3: Prove the result is true for n = k + 1 1 i.e. Prove : 12  32  52    2k  1  k  12k  12k  3 2 3
Proof:
12  32  52    2k  1 2 Proof:  12  32  52    2k  1  2k  1 2 2
12  32  52    2k  1 2 Proof:  12  32  52    2k  1  2k  1 2 2     Sk
12  32  52    2k  1 2 Proof:  12  32  52    2k  1  2k  1 2 2      Sk Tk 1
12  32  52    2k  1 2 Proof:  12  32  52    2k  1  2k  1 2 2      Sk Tk 1 1  k 2k  12k  1  2k  1 2 3
12  32  52    2k  1 2 Proof:  12  32  52    2k  1  2k  1 2 2      Sk Tk 1 1  k 2k  12k  1  2k  1 2 3 1   2k  1 k 2k  1  2k  1 3 
12  32  52    2k  1 2 Proof:  12  32  52    2k  1  2k  1 2 2      Sk Tk 1 1  k 2k  12k  1  2k  1 2 3 1   2k  1 k 2k  1  2k  1 3  1  2k  1k 2k  1  32k  1 3
12  32  52    2k  1 2 Proof:  12  32  52    2k  1  2k  1 2 2      Sk Tk 1 1  k 2k  12k  1  2k  1 2 3 1   2k  1 k 2k  1  2k  1 3  1  2k  1k 2k  1  32k  1 3  2k  12k 2  k  6k  3 1 3  2k  12k 2  5k  3 1 3
12  32  52    2k  1 2 Proof:  12  32  52    2k  1  2k  1 2 2      Sk Tk 1 1  k 2k  12k  1  2k  1 2 3 1   2k  1 k 2k  1  2k  1 3  1  2k  1k 2k  1  32k  1 3  2k  12k 2  k  6k  3 1 3  2k  12k 2  5k  3 1 3 1  2k  1k  12k  3 3
12  32  52    2k  1 2 Proof:  12  32  52    2k  1  2k  1 2 2      Sk Tk 1 1  k 2k  12k  1  2k  1 2 3 1   2k  1 k 2k  1  2k  1 3  1  2k  1k 2k  1  32k  1 3  2k  12k 2  k  6k  3 1 3  2k  12k 2  5k  3 1 3 1  2k  1k  12k  3 3 Hence the result is true for n = k + 1 if it is also true for n = k
Step 4: Since the result is true for n = 1, then the result is true for all positive integral values of n by induction
n 1 n ii    k 1 2k  12k  1 2n  1
n 1 n ii    k 1 2k  12k  1 2n  1 1 1 1 1 n     1 3 3  5 5  7 2n  12n  1 2n  1
n 1 n ii    k 1 2k  12k  1 2n  1 1 1 1 1 n     1 3 3  5 5  7 2n  12n  1 2n  1 Step 1: Prove the result is true for n = 1
n 1 n ii    k 1 2k  12k  1 2n  1 1 1 1 1 n     1 3 3  5 5  7 2n  12n  1 2n  1 Step 1: Prove the result is true for n = 1 1 LHS  1 3 1  3
n 1 n ii    k 1 2k  12k  1 2n  1 1 1 1 1 n     1 3 3  5 5  7 2n  12n  1 2n  1 Step 1: Prove the result is true for n = 1 1 1 LHS  RHS  1 3 2 1 1 1   3 3
n 1 n ii    k 1 2k  12k  1 2n  1 1 1 1 1 n     1 3 3  5 5  7 2n  12n  1 2n  1 Step 1: Prove the result is true for n = 1 1 1 LHS  RHS  1 3 2 1 1 1   3 3  LHS  RHS
n 1 n ii    k 1 2k  12k  1 2n  1 1 1 1 1 n     1 3 3  5 5  7 2n  12n  1 2n  1 Step 1: Prove the result is true for n = 1 1 1 LHS  RHS  1 3 2 1 1 1   3 3  LHS  RHS Hence the result is true for n = 1
n 1 n ii    k 1 2k  12k  1 2n  1 1 1 1 1 n     1 3 3  5 5  7 2n  12n  1 2n  1 Step 1: Prove the result is true for n = 1 1 1 LHS  RHS  1 3 2 1 1 1   3 3  LHS  RHS Hence the result is true for n = 1 Step 2: Assume the result is true for n = k, where k is a positive integer
n 1 n ii    k 1 2k  12k  1 2n  1 1 1 1 1 n     1 3 3  5 5  7 2n  12n  1 2n  1 Step 1: Prove the result is true for n = 1 1 1 LHS  RHS  1 3 2 1 1 1   3 3  LHS  RHS Hence the result is true for n = 1 Step 2: Assume the result is true for n = k, where k is a positive integer 1 1 1 1 k i.e.     1 3 3  5 5  7 2k  12k  1 2k  1
Step 3: Prove the result is true for n = k + 1
Step 3: Prove the result is true for n = k + 1 1 1 1 1 k 1 i.e. Prove :     1 3 3  5 5  7 2k  12k  3 2k  3
Step 3: Prove the result is true for n = k + 1 1 1 1 1 k 1 i.e. Prove :     1 3 3  5 5  7 2k  12k  3 2k  3 Proof:
Step 3: Prove the result is true for n = k + 1 1 1 1 1 k 1 i.e. Prove :     1 3 3  5 5  7 2k  12k  3 2k  3 Proof: 1 1 1 1    1 3 3  5 5  7 2k  12k  3 1 1 1 1 1      1 3 3  5 5  7 2k  12k  1 2k  12k  3
Step 3: Prove the result is true for n = k + 1 1 1 1 1 k 1 i.e. Prove :     1 3 3  5 5  7 2k  12k  3 2k  3 Proof: 1 1 1 1    1 3 3  5 5  7 2k  12k  3 1 1 1 1 1      1 3 3  5 5  7 2k  12k  1 2k  12k  3 k 1   2k  1 2k  12k  3
Step 3: Prove the result is true for n = k + 1 1 1 1 1 k 1 i.e. Prove :     1 3 3  5 5  7 2k  12k  3 2k  3 Proof: 1 1 1 1    1 3 3  5 5  7 2k  12k  3 1 1 1 1 1      1 3 3  5 5  7 2k  12k  1 2k  12k  3 k 1   2k  1 2k  12k  3 k 2k  3  1  2k  12k  3
Step 3: Prove the result is true for n = k + 1 1 1 1 1 k 1 i.e. Prove :     1 3 3  5 5  7 2k  12k  3 2k  3 Proof: 1 1 1 1    1 3 3  5 5  7 2k  12k  3 1 1 1 1 1      1 3 3  5 5  7 2k  12k  1 2k  12k  3 k 1   2k  1 2k  12k  3 k 2k  3  1  2k  12k  3 2k 2  3k  1  2k  12k  3
Step 3: Prove the result is true for n = k + 1 1 1 1 1 k 1 i.e. Prove :     1 3 3  5 5  7 2k  12k  3 2k  3 Proof: 1 1 1 1    1 3 3  5 5  7 2k  12k  3 1 1 1 1 1      1 3 3  5 5  7 2k  12k  1 2k  12k  3 k 1   2k  1 2k  12k  3 k 2k  3  1  2k  12k  3 2k 2  3k  1  2k  12k  3  2k  1k  1 2k  12k  3
 k  1 2k  3
 k  1 2k  3 Hence the result is true for n = k + 1 if it is also true for n = k
 k  1 2k  3 Hence the result is true for n = k + 1 if it is also true for n = k Step 4: Since the result is true for n = 1, then the result is true for all positive integral values of n by induction
 k  1 2k  3 Hence the result is true for n = k + 1 if it is also true for n = k Step 4: Since the result is true for n = 1, then the result is true for all positive integral values of n by induction Exercise 6N; 1 ace etc, 10(polygon), 13