What is mathematical inductive hypothesis? Can anyone give me some examples?

Simple example
4 + 7 + 10 +- - - - + 3n+2= ½(3n² +5n ) prove by inductive

assume it is true ( hypothesis ) and then

If n= n+1 the sum should be ½ [3(n+1)² +5(n+1) prove it

The term of n+1 = 3(n+1)+2

The sum of (n+1) = ½(3n² +5n )+ 3(n+1)+2

½(3n² +5n + 6n+6+4)= ½ [(3n² +6n + 3) + 5n+5 = ½[3(n+1)²+ 5(n+1)]
Qed
the hypothesis is true

Inductive Hypothesis

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Mathematical induction is pretty cool. So, in your first step, you show that the statement is true for the base case, here, you chose n = 1. You plugged in 1 for n, and cool, the statement is true. Now, what you are doing in steps two and three is to show that if you know any arbitrary case is true, the one immediately following it is also true. So, that's why you assume the statement is true for k, then show it is true for the case right after that, k + 1. So, once you finish steps two and three, you have proven that if n = k works, then so does n = k + 1. You have already shown that n = 1 works, so therefore, n = 2 also works. Since you know that n = 2 works, then n = 3 works, and so on, so you have shown that it is true for any natural n. Another one you can prove is this: Prove that 3^n - 1 is divisible by 2.

This is a tough example to start induction with! #1) Since 7^k -1 is divisible by 6, we know that 7^k -1 is a multiple of 6. Algebraically, we may write that as 7^k - 1 = 6r for some integer r. #2) We want to use the fact that 7^k -1 is a multiple of 6 to prove that 7^(k+1) -1 is a multiple of 6. That's what step 3 is all about. Here's my take on step 3. Rewrite 7^(k+1) -1 so that 7^k is visible; anything that is not a multiple of 7^k - 1 should automatically be a multiple of 6. Let's see: 7^(k+1) -1 = 7 * 7^k - 1 (Now the 7^k of the 7^k - 1is exposed!) = 7 * (7^k - 1 + 1) - 1 (adding 0 creatively to completely expose the 7^k -1) = [7 * (7^k - 1) + 7] - 1 (multiplication) = 7 * (7^k -1) + 6. We are assuming that (7^k -1) is divisible by 6. So, the first term 7 * (7^k -1) is also divisible by 6. Clearly, the second term 6 is divisible by 6. Therefore, their sum, which happens to be 7^(k+1) -1 is also divisible by 6. #3-5) By showing that (i) the claim is true for n = 1(the smallest positive integer) as well as (ii) whenever the claim is true for n = k, it is also true for n = k+1, the principle of mathematical induction asserts that the claim is true for all positive integers n. Think of it this way: Since the claim is true for n =1, (ii) implies that it's also true for n = 1 + 1 = 2. Then, since it's true for n = 2, (ii) implies that it's also true for n = 2 + 1 = 3, et cetera. In summary, you need to do two things in an inductive proof. (I) Show the claim is true for the smallest integer n_0(usually n = 0 or 1; if there's a summation, it's the beginning index number). (II) Assuming that the claim is true for n = k (for some k), use this to show that the claim is also true for n = k+1. The Principle of Mathematical Induction then (for free) lets you say that the claim is true for all integers greater than or equal to n_0. --------------------------------------... Here's one example. Show that sum(j=0 to n) 2^j = 2^(n+1) - 1. Proof by induction: Step I (Base Case): Show that the claim is true for n = 0. sum(j=0 to 0) 2^j = 2^0 = 1. 2^(0+1) - 1 = 2-1 = 1. Hence, sum(j=0 to 0) 2^j = 2^(0+1) - 1. Step II (Inductive Step): Assume that the claim is true for n = k. In other words, we assume sum(j=0 to k) 2^j = 2^(k+1) - 1 [just let n = k.] Use this to show that the claim is true for n = k+1. I.e., show that sum(j=0 to k+1) 2^j = 2^(k+2) - 1 [replace n = k+1] To show this, we break the sum on the left into two parts and apply the "inductive hypothesis" (n = k case) sum(j=0 to k+1) 2^j = (first k terms) + (k+1-th term) = [sum(j=0 to k) 2^j] + 2^(k+1) = (2^(k+1) - 1) + 2^(k+1), by n = k case [Now use algebra to make the right hand side look like the (k+1) case!] = 2 * 2^(k+1) - 1, by combining like terms = 2^(k+2) - 1, by a^m * a^n = a^(m+n). So, we have shown that sum(j=0 to k+1) 2^j = 2^(k+2) - 1. This completes this step. Therefore, by the Principle of Mathematical Induction, the claim is true for all integers n >= 0. --------------------------------------... I hope that clarifies matters a bit. Good luck!

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RE:
What is mathematical inductive hypothesis? Can anyone give me some examples?
What is mathematical inductive hypothesis? Can anyone give me some examples?

THX!!!

It is a mathematical model used in Abstract Algebra for proofs.

If you have a formula and wish to prove that it is true, then you must first prove that it holds for n=1. Then, you must prove that it holds for n+1. If both conditions hold, then the proof holds true.

An excellent example is found on the wikipedia page for Mathematical Induction. Scroll down to the "Example" where we use induction to prove
0+1+2+...+n = n(n+1)/2

if others win, i should win too