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Im following

okay

I know, thank you, good notes

so we can rewrite the statement P(n)

ok im starting to understand

|dw:1366641535256:dw|

so we have here a left side and a right side of the statement we are trying to prove

P(1) : |dw:1366641602052:dw|

now the left side is just another way of saying, plug in 1 into the formula 2k+1 , no big deal.

I thought p(1) was 1(1+2)

we want to prove
[2(1) +1] + [2(2) + 1] + [2(3)+1] + ... [2(n)+1] = n(n+1)

oh

yes

because 3 = 2(1) + 1
5 = 2 (2) + 1
7 = 2(3) + 1

|dw:1366641946782:dw|

let's look at the case n=2 |dw:1366642018761:dw|

|dw:1366642104889:dw|

yes, but which number is k because in my problem k isnt named

oh dear i didnt copy correctly

3=3

with me so far?

yeah

using only algebra, basically

so how about we look at the conclusion for a minute

ok, let me do it another way, we can go back to this way later, it might be a little advanced

ok still with me?

yea Im following

we assume that
[2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2) is true

|dw:1366643507200:dw|

so we are just using algebra here

ok

you can add the same number to both sides of an equation, thats a legal algebraic move
dfsd

ok im back , :)

ok

so we added [2(k+1) + 1 ] to both sides of the hypothesis , which was given

hello?

yes did you add it?

|dw:1366644222445:dw|

now the left side is what we want in our conclusion, and we can simplify the right side

okay

Im here but for some reason I keep losing connection

me too

i was lagged

me too

ok did you add both sides

I tried but my teacher said we already solved for what was needed.