help with ANOTHER ruse 4U Q (1 Viewer)

[mreditor16]

so I am doing this q



and I have done all parts except the last one. I have proven true for n=1 and n=2, by using i)

and then assumed true for n=k and n=k+1

and now trying to prove true for n=k+2

and i tried using the assumptions and the provided result, but it is going nowhere.

help anyone?

thanks

 

[mreditor16]

bumpity bump

 

[faisalabdul16]

hmm idk where i would use the result.

If you show (-1)^k+1 / x+k+1 = -T(k,x+1) then it would be done i think.

I am stuck

 

[mreditor16]

hmm idk where i would use the result.

If you show (-1)^k+1 / x+k+1 = -T(k,x+1) then it would be done i think.

I am stuck

oh good, you being stuck makes me feel better.

at least I know I'm not being stupid and not able to get out a easy-ish question....

 

[Machiavelli1]

For n+2:
1. Use result (k+1Cr)=(kCr)+(k+1Cr-1) on LHS
2. Group terms to look like your assumptions
3. Use assumptions and you should have part ii.
I did this Q awhile ago, I think this is how I did it can't remember exactly haha :/

 

[mreditor16]

For n+2:
1. Use result (k+1Cr)=(kCr)+(k+1Cr-1) on LHS
2. Group terms to look like your assumptions
3. Use assumptions and you should have part ii.
I did this Q awhile ago, I think this is how I did it can't remember exactly haha :/

I tried exactly that, but it didn't work. like legit exactly that :O

 

[emilios]

Wait wut why do we prove for n=k+2 ? The condition is n>=1 so wouldn't it make sense for us to use the following steps:
Prove for n=1
Assume n=k is true
Prove for n=k+1

or am i dumb

 

[mreditor16]

Wait wut why do we prove for n=k+2 ? The condition is n>=1 so wouldn't it make sense for us to use the following steps:
Prove for n=1
Assume n=k is true
Prove for n=k+1

or am i dumb

in part one, they asked to prove true for n=1 and n=2. so I took that as strong proof/indication/hint that they want us to prove for n=k+2

 

[Machiavelli1]

I tried exactly that, but it didn't work. like legit exactly that :O

Hmmmmm.... Maybe no need for two assumptions then?

 

[mreditor16]

Hmmmmm.... Maybe no need for two assumptions then?

ruse solutions use two assumptions.

but even still their solutions are an absolute mess and I ceebs posting it because they don't use the assume true for n=k and n=k+1 to then prove true for n=k+2. they do some weird way where they use assumptions for x=x and x=x+1 :/

its just chaotic. argh :/

here it is actually just in case it helps anyone:

 

[Kurosaki]

Ah I see, they're using x as a dummy variable I think it's called.

Edit: Nvm I think im wrong

 

[mreditor16]

Ah I see, they're using x as a dummy variable I think it's called.

huh? care to share?

well using the solutions provided, is it possible to devise a way of doing the question where you assume true for n=k and n=k+1

and then prove true for n=k+2

because that's the way I've been taught for complicated 4u induction qs.

 

[mreditor16]

come on people! we can do this

 

[emilios]

OK I just got it out and checked the solution and my method was the exact same.

First of all forget about making two assumptions, just use one.

Then use the given result to change all the k+1Cr terms to (kCr+kCr-1) terms AND since the first term is k+1C0 just change it to kC0 for uniformity sake i.e. the L.H.S becomes:

kC0/x - (kC1 + kC0)/x+1 + (kC2 + kC1)/x+2 ....

Becomes the grouping shown in the solution

Then you can apply the result of pt ii.

Quite messy but it works out

 

[Collected]

just doing the inductive step (prove true for n=k+1)


LHS=(k+1c0)/x - (k+1c1)/x+1 + .... (-1)^(k+1) (k+1)c(k+1)/(x+k+1)

= kc0/x - (kc1+kc0)/x+1 + (kc2+kc1)/x+2 .... (-1)^(k+1) (kck)/(x+k+1) as (kck=k+1ck+1)

= k!/x(x+1)(x+2)..... - (kc0/x+1 - kc1/x+2 ......) (by assumption)

now from ii we know that T(k,x) - T(k, x+1) = T(k+1, x)

=(k+1)!/x(x+1)...(x+n+1)

=RHS

 

[mreditor16]

OK I just got it out and checked the solution and my method was the exact same.

First of all forget about making two assumptions, just use one.

Then use change all the k+1Cr terms to (kCr+kCr-1) terms AND since the first term is k+1C0 just change it to kC0 for uniformity sake i.e. the L.H.S becomes:

kC0/x - (kC1 + kC0)/x+1 + (kC2 + kC1)/x+2 ....

Becomes the grouping shown in the solution

Then you can apply the result of pt ii.

Quite messy but it works out

okay i'll give it a go. then why the hell do they say using assumptions for x=x and x=x+1 :/

 

[faisalabdul16]

ahh got it now.

This is ugly though...

Probably would have been frustrated in an exam and left it

 

[emilios]

just doing the inductive step (prove true for n=k+1)


LHS=(k+1c0)/x - (k+1c1)/x+1 + .... (-1)^(k+1) (k+1)c(k+1)/(x+k+1)

= kc0/x - (kc1+kc0)/x+1 + (kc2+kc1)/x+2 .... (-1)^(k+1) (kck)/(x+k+1) as (kck=k+1ck+1)

= k!/x(x+1)(x+2)..... - (kc0/x+1 - kc1/x+2 ......) (by assumption)

now from ii we know that T(k,x) - T(k, x+1) = T(k+1, x)

=(k+1)!/x(x+1)...(x+n+1)

=RHS

Yep! You got it.

I think the issue at hand is how to know whether we should use one or two assumptions. The first two parts of the question were quite misleading, but ultimately you know not to use two assumptions since the statement you're proving only uses k, whereas some statements have terms of k and k+1 already within them, if that makes sense (hope i don't confuse you).

 

[Collected]

The ruse solutions say that because they are using the results from part ii (with the t,k - t, k+1) which admittedly is worded pretty badly.

 

[mreditor16]

ahh got it now.

This is ugly though...

Probably would have been frustrated in an exam and left it

same