User talk:Putgeminmouth

Problem
Prove that $$5+10+\cdot\cdot\cdot+5n = {5n(n+1) \over 2}$$

Proof
For convenience let us call Call $$F_n$$ the statement $$5+10+\cdot\cdot\cdot+5n = {5n(n+1) \over 2}$$

So $$F_1~is~5 = {5(1)((1)+1) \over 2}$$

$$F_2~is~5+5(2) = {5(2)((2)+1) \over 2}$$

$$F_{n+1}~is~5+10+\cdot\cdot\cdot+5n+5(n+1) = {5(n+1)((n+1)+1) \over 2}$$

This is just a convenient way of referring to it so we don't have to write it over and over

1) For n=1, we have $$5 = {5(1)((1)+1) \over 2} = {5*2 \over 2} = 5$$ OK

2) We must prove that if $$F_k$$ is true (for some integer k) then $$F_{k+1}$$ must also be true.

Let us then start with $$F_k$$:

$$5+10+\cdot\cdot\cdot+5k = {5k(k+1) \over 2}$$

We can add the same number to both sides and it will still be true, so let us add 5k+1

$$5+10+\cdot\cdot\cdot+5k + 5(k+1)= {5k(k+1) \over 2} + 5(k+1)$$

To add the terms on the right hand side we will put them over a common denominator:

$$5+10+\cdot\cdot\cdot+5k + 5(k+1)= {5k(k+1) \over 2} + {5(k+1)*2 \over 2} = {5k^2+5k+10k+5*2 \over 2} = {5k^2+15k+10 \over 2}$$

$$5+10+\cdot\cdot\cdot+5k + 5(k+1)={5(k^2+3k+2) \over 2} = {5(k+1)(k+2) \over 2} = {5(k+1)(k+1+1) \over 2}= {5(k+1)((k+1)+1) \over 2}$$

Taking just the far left and far right parts of the equation: $$5+10+\cdot\cdot\cdot+5k + 5(k+1)={5(k+1)((k+1)+1) \over 2}$$

But this is the statement of $$F_{k+1}$$! Since we started with $$F_k$$ and ended up with $$F_{k+1}$$ without changing the truth value then we have proven 2)

1) and 2) together make the proof for all integers >= 1

QED