Does anybody know anything about mathematical induction?

Question

The problem says to use mathematical inductions to prove that 2+4+6+...2n=n^2+n holds for all positive integers.

Can somebody explain.

Answer 1

First you must show that this equation is true for n=1 (this is the easy step):
2 × 1 = 1^2 + 1.
Does it check out? Indeed, 2=2.
Now you know the equation holds for at least one value of n. So you suppose that there exists a 'k' such that 2+4+6+...+2k=k^2+k. This is called the induction hypothesis. From this equation, which we know to be true, can we show that:
2+4+6+...+2k+2(k+1) = (k+1)^2 + (k+1) (this is the difficult part)?
Always start with your induction hypothesis:
2+4+6+...+2k=k^2+k
You can add 2(k+1) to both sides:
2+4+6+...+2k + 2(k+1) = k^2 + k + 2(k+1),
... = k^2 + k +2k + 2,
... = (k^2 +2k + 1) + k + 1,
... = (k+1)^2 + (k+1).
Now you say (rather triumphantly) that the induction hypothesis holds for n=k+1 and hence by the Principle of Mathematical Induction, 2+4+6+...+2n = n^2+n for all positive integers n.

Feel fre to email me for any clarifications.

Answer 2

n^2 + n = summation of (2n) for all n element of {1, 2, 3, ... ,n}
n^2 + n = 2(1) + 2(2) + 2(3) + ... + 2n
for all n element of {1, 2, 3, ... ,n}

The inductive processing is used in scientific experimentation. Run the same experiment multiple times and then based on your results draw a conclusion about what will happen each time if the conditions of the experiment are duplicated.

We will demonstrate examples for the conclusion stated above using the positive integers 1, 2, 3 & 99. Based on the results of the trials we can be comfortable that our conclusion is correct. Remember it only takes one exception to prove the conclusion of an inductive process false.

1^2 + 1 = 2(1)
1 + 1 = 2
2 = 2

2^2 + 2 = 2(1) + 2(2)
4 + 2 = 6
6 = 6

3^2 + 3 = 2(1) + 2(2) + 2(3)
9 + 3 = 12
12 = 12

99^2 + 99 = 2(1) + 2(2) + 2(3) + ... + 2(99)
9801 + 99 = 9900
9900 = 9900

our conlcusion is:
n^2 + n = summation of (2n) for all n element of {1, 2, 3, ..,n}

Answer 3

Weak mathematical induction basically amounts to showing that if a statement f(1) holds true, and if whenever f(n) is true, you can show that f(n + 1) is true, then f(m) is true for all m.
In your case, your statement f(1) is 2*1 = 1^2 + 1. This statement is true. Thus, suppose f(n) is true for some value of n. That means 2 + 4 + ... + 2n = n^2 + n for some n. We have already shown that this holds true for at least one value of n (1), so we are not being facetious. We now have to show that this statement implies that 2 + 4 + ... + 2n + 2(n + 1) = (n+1)^2 + (n + 1). Can you do this? If so, you have proven it true for all possible whole numbers, since f(1) true implies f(2) true which implies f(3) true and so on ad infinitum. You are basically writing an infinite amount of proofs once by using symbols in place of proving the statement for each number.

PS. As a hint, note that we already assumed that 2 + 4 + ... + 2n = n^2 + n, so the proof for n + 1 is simplified to showing that n^2 + n + 2(n + 1) = (n + 1)^2 + (n + 1).

Answer 4

This question is synonymous with asking to prove that the sum of the first n integers equals:

1+2+3+...+n = n(n+1)/2

The only change is that both sides were multiplied by 2 and then the n(n+1) term expanded.

The formula I provided above is a classic one shown in any math textbook covering Mathematical Induction, and its proof is often credited to a young Gauss (later famous for other discoveries), and goes something like this:

Suppose you were asked to add up the integers 1 through 100. Gauss (allegedly) realized that adding up pairs of numbers like (1+100), (2+99), (3+98), etc., yielded 50 sums which all added up to 101. Fifty 101's equals 5050, which IS the sum of the first 100 integers.

In this example, we have n=100. And the sum sought did turn out to equal n(n+1)/2. But does this principle hold for ANY value of n? That's what Mathematical Induction seeks to show.

I've included a link to the classic proof, below.