Mathematical Induction?

Okay, proof by induction is first assuming something is true for, say, n, and then showing that it must be true for n+1 as well, which means thereafter it must be true for any n+a, where a is any integer. To illustrate this principle, let's look at n^3 -7n +3, in the case of n = 1.

1^3 - 7(1) + 3 = - 6

which is divisible by 3. Very good. Now, let's assume that

n^3 - 7n + 3 is divisible by 3. Let's see if it's true as well for n+1:

(n+1)^3 -7(n+1) + 3 =

n*3 + 3n^2 + 3n + 1 - 7n - 7 + 3 =

(n*3 - 7n + 3) + (3n^2 + 3n -6)

But we've already said that (n^3 - 7n + 3) is divisible by 3, and (3n^2 + 3n - 6) = 3(n^2 + n 2) is certainly divisible by 3, so that we've proven that it's true for n+1 as well. Thus we've also proved it for any n+a, and the proof by induction is done.

Let P(n) be the statement: 3 divides n^3 - 7n + 3.

Base case: check that P(0) is true...that is 3 divides 0^3 - 7(0) + 3. Ok it does.

Induction: Assume P(k) and try to show that P(k) implies P(k+1).

P(k) is true means 3 divides k^3 - 7k + 3.

Now we look at (k+1)^3 - 7(k+1) + 3. That equals k^3 + 3k^2 - 4k - 3 = (k^3 -7k + 3) + (3k^2 + 3k - 6) = (k^3 -7k + 3) + 3(k^2 + k - 2).

We know 3 divides k^3 - 7k + 3, and there's an obvious factor of 3 in the second piece so 3 divides it too. Therefore 3 divides the whole thing. Since we've proven P(0) and that P(k) implies P(k+1), by induction we know P(n) is true for all non-negative integers "n."

First show this is true for n = 1

1^3 - 7(1) + 3 = -3 (divisible by three)

2nd assume that it is true for any value n.

3rd, show that it is true for n + 1

(n+1)^3 - 7(n + 1) + 3 = n^3 + 3n^2 - 4n - 3

If we add 3n (which is divisible by 3) to the original expression (also divisible by 3), we get

n^3 - 4n - 3, which we know is divisible by 3 because any two terms added together that are divisible by 3, is also divisible by 3.

and 3n^2 is obviously divisible by 3, so the expression...

n^3 + 3n^2 - 4n - 3 is divisible by three.

This proves that n^3 - 7n + 3 is divisible by 3 for all non-negative integers.

you want 2 issues: a base case and an "inductive leap of religion". then you definitely teach both. in this actual party, your base case will be S(a million), and your "leap of religion" will be that even if it really is actual for S(n), then it really is likewise actual for S(n+a million). S(a million) = a million^2 = a million * 2 * 3 / 6 = a million. Sp. actual for S(n), teach for S(n + a million). purpose: teach that S(n+a million) = (n + a million)(n + 2)(2n + 3) / 6 = (n^2 + 3n + 2)(2n + 3) / 6 = (2n^3 + 9n^2 + 13n + 6) / 6 start up with the definition of S(n+a million): S(n+a million) = S(n) + (n+a million)^2 = S(n) + n^2 + 2n + a million and then replace the equation for S(n), utilizing the inductive hypothesis: S(n+a million) = n(n + a million)(2n + a million) / 6 + n^2 + 2n + a million = (2n^3 + 3n^2 + n) / 6 + n^2 + 2n + a million = ((2n^3 + 3n^2 + n) + (6n^2 + 12n + 6)) / 6 = (2n^3 + 9n^2 + 13n + 6) / 6 with assistance from the way, you should verify you recognize this information, and then attempt to video demonstrate it to something else.