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