Binomial Coefficient Comparison

[harv93]

Hi All,

Im new to this forum and have never really thought about joining up to a maths forum, i have been part of programming forum for some years however I am struggling on some maths and have nowhere to turn. The question is relatively simple i just cant seem to get my head around this.

Could somebody walk me through there process and the steps they would take for this, I just need be shown this once and i'm 100% sure I could complete any similar question. I haven't done anything like this for around 6 years.

Question:


Thanks in advance,

 

[pka]

Question:
View attachment 5029

One can answer the first part in any number of ways. Each answer would be equally valid.

However, there is only one class of answers that helps with the second part. In fact one must already know the binomial expansion theorem in order to answer that part. It ia dumb way to ask this important question.

The binomial expansion theorem: \(\displaystyle {\displaystyle{\left( {x + y} \right)^n} = \sum\limits_{k = 0}^n {\binom{n}{k}{x^k}{y^{n - k}}}} \)

Note that \(\displaystyle \left( {1+1} \right)^n=2^n~.\) So let \(\displaystyle x=~?~\&~y=~?\)

 

[harv93]

One can answer the first part in any number of ways. Each answer would be equally valid.

However, there is only one class of answers that helps with the second part. In fact one must already know the binomial expansion theorem in order to answer that part. It ia dumb way to ask this important question.

The binomial expansion theorem: \(\displaystyle {\displaystyle{\left( {x + y} \right)^n} = \sum\limits_{k = 0}^n {\binom{n}{k}{x^k}{y^{n - k}}}} \)

Note that \(\displaystyle \left( {1+1} \right)^n=2^n~.\) So let \(\displaystyle x=~?~\&~y=~?\)

Hi PKA,

Thanks for your swift reply, i am still unsure of where to go , I did binomial expansion back in school years ago for example (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3..

I just cant get my head around the question , maybe I am completely over looking it? I cannot even write a sentence to relate it to the set....

 

[Ishuda]

...It ia dumb way to ask this important question. ...[/tex]

Thank you! I thought I was maybe feeling a little perverse today when I had the same thought. Glad to see it somewhere else.

 

[pka]

i am still unsure of where to go , I did binomial expansion back in school years ago for example \(\displaystyle (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3\)

\(\displaystyle (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3\)

\(\displaystyle (1+1)^3 = 1^3 + 3\cdot 1^2\cdot 1 + 3\cdot 1\cdot 1^2 + 1^3=8=2^3 \)

 

[harv93]

Can anybody else give me some input,

thanks in advance

 

[Ishuda]

Can anybody else give me some input,

thanks in advance

I think pka covered it nicely. I would have said basically the same thing including the "It is a dumb way to ask this important question". Look again at the formula for the binomial expansion as posted by pka and substitute 1 for x and y as suggested. You should have this formula (the binomial expansion) memorized - at least for the time you are doing mathematics on any regular basis.