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lol that's from reddit

\[\sum_{1}^{i=1}(1)2^{(1)}=\]

Whats from reddit?

So above, on the left i have 2

on the right....

It's the sum up to n + 1, not n

\[(1)2^{((1)+2)}+2 = 2^{3}+2 = 10\]

@ strobe, so i am supposed to plug in a 2 instead of a 1?

It's the sum of the LHS subbed with 1 and 2

LHS?

left hand side

Yah

ok so the RHS is right

\[\sum_{i = 2}^{2} = 2(2)^{2} = 8\]

so add that to the first summation and i get 10 on the LHS

so 10 = 10 is true

Is that correct?

Yes

Ok, thanks, ill see if I can finish the rest from there :)

Good luck!

after step one (show p(1) is true:
step 2 is assume p(k) is true and then show p(k) implies p(k+1)

Cant quite see how i can make that happen at the moment, heres what I have:

\[p(k) : \sum_{i=1}^{k+1}i2^{i} = k2^{k+2}+2\]

\[p(k+1): \sum_{i=1}^{k+2}i2^{i}=(k+1)2^{k+3}+2\]

RHS simplifies to \[k2^{k+3}+2^{k+3}+2\]

So now I need to add something to p(k) to make it match p(k+1)

and if after adding that to p(k) the RHS of p(k) = the RHS of p(k+1), that is sufficient proof

\[p(k)+(k+1)2^{k+1}: \sum_{i=1}^{k+1}i2^{i}+(k+1)2^{k+1}=k2^{k+2}+2+(k+1)2^{k+1}\]

This is so freakin complicated

So first I should calculate the right hand side of p(k+1)?

or of p(k)

the LHS of p(k+1)

ok so,

RHS of p(k+1): \[(k+1)2^{(k+1)+2}+2\]

isnt much calculation to do really, just plug in

Then I want to make LHS of P(k+1) the same as that:

RHS of p(k+1): \[\sum_{i=1}^{(k+1)+1}i2^{i}\]

*LHS rather

Yup, so that equals the LHS of p(k) + something, find that something

So I should expand the LHS of p(k+1) first?

Expand so it looks like LHS of p(k) + something

I would need to add (k+1)2^(k+1) to LHS of p(k) for them to be equal

I just have to show that...

\[(p(k)LHS)+(k+1)2 ^{k+1}-> (k2^{k+2}+2)+(k+1)2 ^{k+1}\]

The substitution into (p(k)LHS) is p(k)RHS

Now I need to make that = the RHS or p(k+1)?

If so, that's the method I am familiar with

Yup, you'll find it all factorises nicely

I'll try it here

so im trying to make \[(k2^{k+2}+2)+(k+1)2^{k+1} = (k+1)2^{(k+1)+2}+2 \]

expanding everything out isnt the way to go?

yes

it is the way to go?

hmm

p(k): \[k2^{k}4+2^{k}2+4\]

Im not seeing it :/

Clearly, im totally lost

Thats what i tried to show

Did i make an arithmetic error somewhere or is my methodology wrong or what

\[\sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)w^{m+2}\]

ok

so now I need to set that equal to the right hand side of m+1

or make it equal that at least

I'm still working on it. we want the right hand side equal to what the formula says if we use n= m+1

RHS of m+1:\[(m+1)2^{m+3}+2 \]

right we need to make that equal what i just typed up right?

If so, that is this: http://imgur.com/P8Tnf

yes
and I believe they are.

Just with m

isnt it or am i missing something?

we have m+2 now, you have k+1

very close, but not close enough!

ok i must have messed that up somewhere

Do you see how to get the "new" version? (with m+2), it's the next term we are adding, with i= (m+2)

yea, so confusing!

multiply out (m+2) 2^(m+2)

making it m made it less tho

lets see

we get
\[ 2+ m 2^{m+2}+m2^{m+2} + 2^{m+3} \]

on LHS right

thats not the same tho

thats \[2+2m ^{m+2}+2^{m+3}\]

on the LHS

add the two m2^(m+2) to get m*2*2^(m+2) = m*2^(m+3)

RHS is \[2+m2^{m+3}+2^{m+3}+2\]

Ah yes

Now I see it :)

you have an extra 2 in there. But factor out 2^(m+3) to get
\[ 2+ (m+1) 2^{m+3} \]

Let me try and write this out and think it out

hopefully fully see where i bliped up

ok have fun

Thanks so much

Will post here when done

http://imgur.com/BYB4E