writing this sum in binomial form: sum, k=1 to 2n, of 2n-choose-k * 2^{-2k} * 3^{2n-k}

[Qwertyuiop[]]

[imath]\sum_{k=1}^{2n}\binom{2n}{k} 2^{-2k} 3^{2n-k}[/imath], i want to express this sum in binomial form but the problem is instead of (n choose k), i have (2n choose k) and the powers of coefficients are not n-k and k. Using the laws of indices i can factor out 2^-2 from 2^(-2k) to get [imath]2^{-2}\sum_{k=1}^{2n}\binom{2n}{k} 2^{k} 3^{2n-k}[/imath] But i can't rewrite 3^(2n-k) to 3^(n-k), I just need a hint.

 

[topsquark]

[imath]\sum_{k=1}^{2n}\binom{2n}{k} 2^{-2k} 3^{2n-k}[/imath], i want to express this sum in binomial form but the problem is instead of (n choose k), i have (2n choose k) and the powers of coefficients are not n-k and k. Using the laws of indices i can factor out 2^-2 from 2^(-2k) to get [imath]2^{-2}\sum_{k=1}^{2n}\binom{2n}{k} 2^{k} 3^{2n-k}[/imath] But i can't rewrite 3^(2n-k) to 3^(n-k), I just need a hint.

Are you sure that sum isn't [imath]\displaystyle \sum_{k = \red{\textbf{0}}}^{2n}[/imath]? If it does start at 1, then you need to add and subtract an appropriate term to do this.

Actually, you can't factor like that. [imath]2^{-2k} \neq 2^{-2} 2^k[/imath].

Your whole problem is to rewrite the [imath]2^{-2k}[/imath] as something like [imath]a^k[/imath]. Then
[imath]\displaystyle \sum_{k = 0}^{2n} {2n \choose k} a^k 3^{2n - k}= (a + 3)^{2n}[/imath]

So if [imath]2^{-2k} = a^k[/imath], what's a?

-Dan

 

[blamocur]

the problem is instead of (n choose k), i have (2n choose k)

Why is this a problem? You can use [imath]2n[/imath] instead of [imath]n[/imath]. The only "problem" is [imath]2^{-2k}[/imath], but this can easily converted to [imath]u^k[/imath], just figure out what appropriate [imath]u[/imath] is in this case.

 

[Qwertyuiop[]]

Are you sure that sum isn't [imath]\displaystyle \sum_{k = \red{\textbf{0}}}^{2n}[/imath]? If it does start at 1, then you need to add and subtract an appropriate term to do this.

Actually, you can't factor like that. [imath]2^{-2k} \neq 2^{-2} 2^k[/imath].

Your whole problem is to rewrite the [imath]2^{-2k}[/imath] as something like [imath]a^k[/imath]. Then
[imath]\displaystyle \sum_{k = 0}^{2n} {2n \choose k} a^k 3^{2n - k}= (a + 3)^{2n}[/imath]

So if [imath]2^{-2k} = a^k[/imath], what's a?

-Dan

no the sum starts at k=1. a=1/4 because 2^(-2k)=2^((-2)k) which is (1/4)^k so this becomes (1/4 + 3)^n and subtracting k=1 gives the final expression [imath](\frac{1}{4}+3)^{n} - \frac{3^{2n-1}}{4}[/imath]. I have a question regarding the binomial formula, does it matter what we choose for a and b? Can we choose a = 3 and b=1/4 ? Because the end result is the same.

 

[Qwertyuiop[]]

Why is this a problem? You can use [imath]2n[/imath] instead of [imath]n[/imath]. The only "problem" is [imath]2^{-2k}[/imath], but this can easily converted to S[imath]u^k[/imath], just figure out what appropriate [imath]u[/imath] is in this case.

So it doesn't have to be (n k) if you want to use binomial formula? I was looking for ways to express (2n k) as (n k ) . But we do need the powers of a and b to match with (n k) ?

 

[Dr.Peterson]

So it doesn't have to be (n k) if you want to use binomial formula? I was looking for ways to express (2n k) as (n k ) . But we do need the powers of a and b to match with (n k) ?

At some point, you may want to make a substitution, such as letting m = 2n, so it looks more like what you want. Never focus on the variables used in a formula; focus on the form and relationships.

Similarly, at some point you will probably want to make some adjustment in the index, which could either be a substitution, or (more likely) just adding or removing terms at the start.

In general, do something, write out the results, and then either decide on the next thing to do, or decide that what you did went in the wrong direction, and back up.

a=1/4 because 2^(-2k)=2^((-2)k) which is (1/4)^k so this becomes (1/4 + 3)^n and subtracting k=1 gives the final expression [imath](\frac{1}{4}+3)^{n} - \frac{3^{2n-1}}{4}[/imath]. I have a question regarding the binomial formula, does it matter what we choose for a and b? Can we choose a = 3 and b=1/4 ? Because the end result is the same.

You've taken a step here; now compare what you have with the summation, and think about what is wrong. Should that exponent really be n, for example? And is it really k=1 that you want to subtract?

 

[topsquark]

no the sum starts at k=1

Then you will need to add and subtract the k = 0 term for the binomial expansion in order to do this.

-Dan

 

[Steven G]

2n is a number just like m is. Just make the substitution m=2n if that will make you happy, ok?

 

[Qwertyuiop[]]

At some point, you may want to make a substitution, such as letting m = 2n, so it looks more like what you want. Never focus on the variables used in a formula; focus on the form and relationships.

Similarly, at some point you will probably want to make some adjustment in the index, which could either be a substitution, or (more likely) just adding or removing terms at the start.

In general, do something, write out the results, and then either decide on the next thing to do, or decide that what you did went in the wrong direction, and back up.

You've taken a step here; now compare what you have with the summation, and think about what is wrong. Should that exponent really be n, for example? And is it really k=1 that you want to subtract?

Oh, i understand now, the binomial formula adds up numbers from k=0 and because we are starting at k=1 , we need to subtract k=0, and the exponent should be 2n, correct?