Help please!
for the given statement Pn, write the statements P1, Pk, and Pk+1. 2+4+6...+2n=n(n+1)
induction
Yes I am aware of what type of problem it is.
well
Well I do not know what to do to get them into the different statements.
hmm Suppose the equation is true for n = k Just replace n by k. 2 + 4 + 6 + ... + 2k = k ( k + 1) . . . . After you replace k by k+1, you get : 2 + 4 + 6 + ... + 2 × (k + 1) = k+1 ( k + 1 + 1) 2 + 4 + 6 + ... + 2 × ( k + 1) = k+1 ( k + 2) 2 + 4 + 6 + ... + 2 × ( k + 1) = ( k + 1 ) × ( k + 2) Add 2 × ( k + 1) to both sides of the hypothesis 2 + 4 + 6 + ... + 2k + 2 × ( k + 1) = k ( k + 1) + 2 × ( k + 1) = k2 + k + 2k + 2 = k2 + 3k + 2 Since 2 = 1 × 2 and 1 + 2 = 3, k2 + 3k + 2 = ( k + 1) × ( k + 2) Therefore, 2 + 4 + 6 + ... + 2k + 2 × ( k + 1) = ( k + 1) × ( k + 2) and the proof by mathematical induction
So you just replace n with whatever the value by P is? so like n=1, n=k, n=k+1? And then solve?
yes in induction u first prove for n=1 then n=k then n=k+1
Okay and how do you get the value of the left side? you plug 1 into the n spot like 2n you would get 2? because I've watched this video: http://www.youtube.com/watch?v=IFqna5F0kW8 and I do not get how to get the left side value.
LOL for n=1 n(n+1) = 1(1+1)=2 n is the number of terms on left side 1st term that means 2
Okay so for my problem there are 3 terms or do you count 2n?
rephrase it pls?
For my problem, the left side is 2+4+6+...+2n. Are there 3 terms or do you count 2n as a term on the left side?
nope let me explain for n=1 its 2(1) = for n(2) its 2(2)=4 and 2n is written for nth term
sorry its for n=2
so for nth term its 2*n
Okay so if I solve it for n=1 it would look like 2=1(1+1) 2=1(2) 2=2
how are you suppossed to solve for k?
Just replace n by k. 2 + 4 + 6 + ... + 2k = k ( k + 1)
i told u the process check it
I actually think you did something incorrect... I think it should be 2+4+6+...+2k+2(k+1)=k+1(k+1+1)
that's for k+1
i have solved correctly
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