# Thread: Some help with simple "how to prove thing"

1. ## Some help with simple "how to prove thing"

Hello everyone!
to start with I am new to this forum and I come from sweden so I am not that good in english mathematic terms. But I will try my best. Also I am going in highschool and not college/university but sinse the course I am taking is named "Descrete mathe" I thought it would fit quiet good here. Anyway, I think this is very easy but I am in year 1 in highschool so for me it's hard.

I need to prove that n = 1,2,3 ...
is working for
a) 2 + 4 +...+ 2n = n(n+1)
b) 1 + 2 + 4 +...+ 2^n = 2^(n+1)-1

I cant even figure out the A..
n=1 VL=2, HL=1(1+1)=2, VL=HL

n=p,
2 + 4 +...+ 2p = p(p+1)

n=p+1,
2 + 4 +...+ 2p + (2p+1) = (p+1)(p+2)
p(p+1)+(2p+1) = (p+1)(p+2)
p^2+p+2p+1 = p^2+p+2p+2

Now why is that wrong, I am doing something wrong..

2. Nevermind I did discovere the problem with some help, i have put 2p inside the (2p+1) when it should be 2(p+1). Now it is just b) left!

3. Hello, Mquis!

Welcome aboard!

It looks like you're using Inductive Proofs.

a) Prove: .$\displaystyle 2 + 4 + 6 + \cdots...+ 2n \:=\:n(n+1)$

Verify $\displaystyle S(1)\!:\;\;2 \:=\:1(1+1) \quad\Rightarrow\quad 2 \:=\:2$ . . . true!

Assume $\displaystyle S(k)\!:\;\;2 + 4 + 6 + \cdots + 2k \:=\:k(k+1)$

Add $\displaystyle 2(k+1)$ to both sides: . $\displaystyle 2 + 4 + 6 + \cdots + 2k + 2(k+1) \:=\:k(k+1) + 2(k+1)$

. . The right side is: .$\displaystyle k^2 + k + 2k + 2 \:=\:k^2+3k + 2 \:=\:(k+1)(k+2)$

The equation becomes: .$\displaystyle 2 + 4 + 6 + \cdots + 2(k+1) \:=\:(k+1)(k+2)$

This is $\displaystyle S(k+1)$ . . . The inductive proof is complete.

b) Prove: .$\displaystyle 1 + 2 + 4 + \cdots + 2^n \:= \:2^{n+1}-1$

Verify $\displaystyle S(1)\!:\;\;1 +2\;=\;2^2-1\quad\Rightarrow\quad 3 \:=\:3$ . . . true!

Assume $\displaystyle S(k)\!:\;\;1 + 2 + 4 + \cdots + 2^k \:=\:2^{k+1} - 1$

Add $\displaystyle 2^{k+1}$ to both sides: .$\displaystyle 1 + 2 + 4 + \cdots + 2^k + 2^{k+1} \;=\;2^{k+1}-1 + 2^{k+1}$

. . The right side is: .$\displaystyle 2\cdot2^{k+1} - 1 \;=\;2^{k+2}-1$

The equation becomes: .$\displaystyle 1 + 2 + 4 + \cdots + 2^{k+1} \;=\;2^{k+2}-1$

This is $\displaystyle S(k+1)$ . . . The inductive proof is complete.

Edit: I see that you figured it out . . . Good for you!
.

4. I just wonder 2 things.

1.How can you say that you have prooved it... dont the sides have to be exactly? Even if that is understood.

2.How do you write codes like that?

Edit; 3. Did I post this in the right section?