small arithmithic problem

easyontheeye

hey,

im doing maths as part of a course im doing, its 6 years since i was in college and it seems ive forgotten every basic rule going! what are the steps involved in achieving this ? its just a small step in a proof of induction.


n(n+1)/2+(n+1) = (n+1)(n/2 +1)


ive attached an image which is easier to read...
thanks!

Pherekydes

easyontheeye wrote: »

hey,

im doing maths as part of a course im doing, its 6 years since i was in college and it seems ive forgotten every basic rule going! what are the steps involved in achieving this ? its just a small step in a proof of induction.


n(n+2)/2+(n+1) = (n+1)(n/2 +1)


ive attached an image which is easier to read...
thanks!

Multiply them out!

easyontheeye

yeah done that,

n(n+1)/2+(n+1) ...

= (n^2 +n) /2 + (n+1)


what now? i dont see how you remove the (n+1) and it becomes a multiplier

Michael Collins

easyontheeye wrote: »

yeah done that,

n(n+1)/2+(n+1) ...

= (n^2 +n) /2 + (n+1)


what now? i dont see how you remove the (n+1) and it becomes a multiplier

Now combine them all into a single ratio to get: quadratic / 2

Now factor out this quadratic.

Then divide the 2 into one of the resulting multipliers.

L'prof

easyontheeye wrote: »

hey,

im doing maths as part of a course im doing, its 6 years since i was in college and it seems ive forgotten every basic rule going! what are the steps involved in achieving this ? its just a small step in a proof of induction.


n(n+1)/2+(n+1) = (n+1)(n/2 +1)


ive attached an image which is easier to read...
thanks!

n(n+1)/2+(n+1) = (n+1)(n/2 +1)

=> (n^2 + n)/2 + (n+1) = (n^2 + n)/2 + (n+1)

You just multiply the left and right hand side out!

Michael Collins

jasonorr wrote: »

n(n+1)/2+(n+1) = (n+1)(n/2 +1)

=> (n^2 + n)/2 + (n+1) = (n^2 + n)/2 + (n+1)

You just multiply the left and right hand side out!

For most things that's OK, but you have to be careful when manipulating both sides of an equation like that when you're trying to prove they are equal. Each step has to have forward and reverse implication.

Like it's easy to start off with 5 = 6 (which is obviously not true)

Multiply both sides by zero 0 = 0 (it might not be obvious that this is what you did)

Then add 3 say to boths sides ending up with 3 = 3 which is of course true, but the original "equation" wasn't.

Anyway, rant over.

L'prof

Michael Collins wrote: »

For most things that's OK, but you have to be careful when manipulating both sides of an equation like that when you're trying to prove they are equal. Each step has to have forward and reverse implication.

Like it's easy to start off with 5 = 6 (which is obviously not true)

Multiply both sides by zero 0 = 0 (it might not be obvious that this is what you did)

Then add 3 say to boths sides ending up with 3 = 3 which is of course true, but the original "equation" wasn't.

Anyway, rant over.

But if n = 0, then 1=1,
if n+1 = 0, then 0=0

easyontheeye

ok i welcome all the feedback,

but when i manipulate the left side i come up with,



n^2/2 + n/2 +n +1

i still cant get to the next step which is (n+1)(n/2 +1)

(i clearly forget the basic rules around manipulating equations)

Thoie

I haven't done this in years, so correct me if I'm wrong.

I multiplied the first line by 2 getting:

n(n+1) + 2(n+1) = 2(n+1)(n/2 +1) ==>

n^2 +n +2n +2 = (n+1)(n+2) ==>

n^2 + 3n +2 = n^2 +2n +n + 2 ==>

n^2 + 3n + 2 = n^2 +3n +2. QED.



To mulitply (A+1) by (B+2) (I'm using different letters so you can see what's happening)..

(A x + (A x 2) + (1 x + (1 x 2) ==> AB + 2A + B + 2.

If we replace As and Bs with ns, we get

nn + 2n + n + 2 ==> n^2 + 3n + 2.

Was that the bit giving trouble?

easyontheeye

Thoie wrote: »

I haven't done this in years, so correct me if I'm wrong.

I multiplied the first line by 2 getting:

n(n+1) + 2(n+1) = 2(n+1)(n/2 +1) ==>

n^2 +n +2n +2 = (n+1)(n+2) ==>

n^2 + 3n +2 = n^2 +2n +n + 2 ==>

n^2 + 3n + 2 = n^2 +3n +2. QED.



To mulitply (A+1) by (B+2) (I'm using different letters so you can see what's happening)..

(A x + (A x 2) + (1 x + (1 x 2) ==> AB + 2A + B + 2.

If we replace As and Bs with ns, we get

nn + 2n + n + 2 ==> n^2 + 3n + 2.


Was that the bit giving trouble?

that did help although
n^2 + 3n + 2 does not equal (n+1)(n/2 +1)

i assume i have to divide it back again by 2?

Michael Collins

easyontheeye wrote: »

ok i welcome all the feedback,

but when i manipulate the left side i come up with,



n^2/2 + n/2 +n +1

i still cant get to the next step which is (n+1)(n/2 +1)

(i clearly forget the basic rules around manipulating equations)

So this is a quadratic function which you can factor. First of all, let's get rid of the 1/2 coefficients like so:

n^2/2 + n/2 +n +1

[n^2 + n + 2 (n + 1)]/2 ...adding fractions

[n^2 + n + 2n + 2]/2

[n^2 + 3n + 2]/2 ...the numerator is a quadratic which can be factored to get

[(n + 1)(n+2)]/2

Now dividing 2 into the second multiplier to get:

(n+1)(n/2 + 1)

which equals the other side

easyontheeye

brilliant factorising!! thats what i forgot!

one last question, why only divide 2 into the second multiplier?



edit:nevermind i see it now

Thoie

easyontheeye wrote: »

that did help although
n^2 + 3n + 2 does not equal (n+1)(n/2 +1)

i assume i have to divide it back again by 2?

You forgot to multiply both sides by 2?

n^2 + 3n +2 = 2(n+1)(n/2 +1)


So what I did was use the 2 to cancel out the n/2, leading to

n^2 + 3n +2 = (n+1)(n+2)

easyontheeye

Thoie wrote: »

You forgot to multiply both sides by 2?

n^2 + 3n +2 = 2(n+1)(n/2 +1)


So what I did was use the 2 to cancel out the n/2, leading to

n^2 + 3n +2 = (n+1)(n+2)

well what i was trying to achieve is how can manipulate the left side to equal the right side and the steps in between.
so the end result should be (n+1)(n/2 +1)

Cokehead Mother

All you really needed to know for this is that xy + xz = x(y + z). [x is (n+1) in this case and y are z are n/2 and 1 respectively]

How did this thread get so complicated?

easyontheeye

Cokehead Mother wrote: »

All you really needed to know for this is that xy + xz = x(y + z). [x is (n+1) in this case and y are z are n/2 and 1 respectively]

How did this thread get so complicated?

my fault

Thoie

Cokehead Mother wrote: »

How did this thread get so complicated?

Because some of us are dragging 20 year old information from the back of our skulls, and didn't understand the original question

LeixlipRed

You're all banned for making my head hurt!!!

easyontheeye

you may as well ban me now because i think im going to cause alot more headaches around here. .. :pac:

Delphi91

n(n+1)/2+(n+1) = (n+1)(n/2 +1)

Concentrate on the left hand side only...

easyontheeye

nice and neat, i like it.

swiss

Its been so long since I did a problem like this I decided to give it a go. The solution given is about as clear as I can make it.

Thoie

Delphi91 wrote: »

n(n+1)/2+(n+1) = (n+1)(n/2 +1)

Concentrate on the left hand side only...

I had problems seeing how you got from line 2 to line 3 - could you spell it out a bit more? If you don't have time, no worries - it's just got my brain fired up now

MathsManiac

Thoie wrote: »

I had problems seeing how you got from line 2 to line 3 - could you spell it out a bit more? If you don't have time, no worries - it's just got my brain fired up now

It's just the same concept that Cokehead pointed out earlier: (n+1) is a common factor in the two terms.

MathsManiac

...or to explain further: algebraic manipulation is often about seeing a familiar structure in an unfamiliar expression: factorising xy+xz as x(y+z) is the same as saying: dog*cat + dog*budgie = dog*(cat+budgie),

or indeed, in this case:

(n+1)*dog + (n+1)*cat = (n+1)*[dog + cat]

On some other occasion, you might be doing trigonometry, and have to be able to see that sin(x)*cos(x) + cos(x)*tan(x) = cos(x)*[sin(x)+tan(x)].

These are all applying precisely the same factorisation principle, and recognising such strucures is the key to becoming competent at algebraic manipulation.

Does that help?

easyontheeye

you either see it or you dont...and i most cases i dont

Thoie

I see it now. For some reason when I saw it the first time I had some mad notion that we were dividing by n/2 A closer look has helped!

Delphi91

easyontheeye wrote: »

you either see it or you dont...and i most cases i dont

The more you do/practice similar problems, the more you get to see these things.