Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

C2 help!

I can't seem to work out person 8i)

I have got up to

Unparseable latex formula:

\drac{n(n-1)(n-2)} {6}=2[\dfrac{n(n-1)(n-2)(n-3)}{24}]

Power of a^(n-3) on x^3 term and a^(n-4) term

Not sure what to do from here.

Thanks in advance :smile:

:smile:

(edited 9 years ago)

C2- Binomial expansion.jpg28.5KB

Original post by Super199

I can't seem to work out person 8i)

I have got up to

Unparseable latex formula:

\drac{n(n-1)(n-2)}{6}=2[\dfrac{n(n-1)(n-2)(n-3)}{24}]

Power of a^(n-3) on x^3 term and a^(n-4) term

Not sure what to do from here.

Thanks in advance :smile:

:smile:

Can you think logically rather than just using an arbitrary formula?
What would seem logical here?

Original post by Super199

...

\dfrac{n(n-1)(n-2)}{6} = 2 \left( \dfrac{n(n-1)(n-2)(n-3)}{24} \right)

You've formed the equality concerning the coefficients of

x^3

and

x^4

. All that's left to do is solve for

n

, taking into account that

n>4

.

(edited 9 years ago)

OP

Original post by m4ths/maths247

Can you think logically rather than just using an arbitrary formula?
What would seem logical here?

I'm not sure? This is the way I have been taught :redface:

:redface:

OP

Original post by Khallil

\dfrac{n(n-1)(n-2)}{6} = 2 \left( \dfrac{n(n-1)(n-2)(n-3)}{24} \right)

You've formed the equality concerning the coefficients of

x^3

and

x^4

. All that's left to do is solve for

n

, taking into account that

n>4

.

That's the part where I was stuck. :confused:

:confused:

Original post by Super199

...

See if you can get it into the form of

f(n) = 0

and factorise

\dfrac{n(n-1)(n-2)}{6}

out of

f(n)

.

Original post by Super199

That's the part where I was stuck. :confused:

:confused:

This is the issue when just plugging numbers into a formula.
Do you have common factors on both sides of your equation you can (a) cancel of (b) factor ?

OP

Original post by Khallil

See if you can get it into the form of

f(n) = 0

and factorise

\dfrac{n(n-1)(n-2)}{6}

out of

f(n)

.

Original post by m4ths/maths247

This is the issue when just plugging numbers into a formula.
Do you have common factors on both sides of your equation you can (a) cancel of (b) factor ?

Got there in the end. Thanks :smile:

:smile:

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