# Converting Sigma notation...

by Pranav-Arora
Tags: converting, notation, sigma
 P: 3,807 1. The problem statement, all variables and given/known data (Not a homework question) Hi!! I have been encountering problems in Binomial Theorem which includes converting the sigma notation. Like $$\sum_{k=0}^n \frac{n!}{(n-k)!k!} a^kb^{n-k}=(a+b)^n$$ I got many questions in my exam of this type with four options. One of them was:- $$\sum_{k=1}^{n} {}^nC_k.3^k$$ I substituted the value of n and was able to figure out the correct option. But as i said there were many questions, so it took a lot of time. Is there any easier way to do that? 2. Relevant equations 3. The attempt at a solution
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P: 3,562
 Quote by Pranav-Arora 1. The problem statement, all variables and given/known data (Not a homework question) Hi!! I have been encountering problems in Binomial Theorem which includes converting the sigma notation. Like $$\sum_{k=0}^n \frac{n!}{(n-k)!k!}=(a+b)^n$$
The binomial theorem is:

$$\sum_{k=0}^n \frac{n!}{(n-k)!k!}a^kb^{n-k}=(a+b)^n$$

So this would assume that a=b=1. I'm not sure if this was a question or you just incorrectly wrote down the binomial expansion though...

 Quote by Pranav-Arora I got many questions in my exam of this type with four options. One of them was:- $$\sum_{k=1}^{n} nC_k.3^k$$ I substituted the value of n and was able to figure out the correct option. But as i said there were many questions, so it took a lot of time. Is there any easier way to do that?
I'm still not quite getting it. That is just an expression, but what is the actual question?
P: 3,807
 Quote by Mentallic The binomial theorem is: $$\sum_{k=0}^n \frac{n!}{(n-k)!k!}a^kb^{n-k}=(a+b)^n$$ So this would assume that a=b=1. I'm not sure if this was a question or you just incorrectly wrote down the binomial expansion though...
Sorry!! I incorrectly wrote down the binomial expansion.

 Quote by Mentallic I'm still not quite getting it. That is just an expression, but what is the actual question?
Like the binomial expansion can be written to (a+b)n, i want to write the given expression in the form as we compress the sigma notation to (a+b)n.
(Would you please tell me how to make the "n" before "C" in Superscript?)

I hope you get it now.

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P: 3,562
Converting Sigma notation...

 Quote by Pranav-Arora Like the binomial expansion can be written to (a+b)n, i want to write the given expression in the form as we compress the sigma notation to (a+b)n. (Would you please tell me how to make the "n" before "C" in Superscript?) I hope you get it now.
Oh ok I see, well then since we have the binomial expansion involves both ak and bn-k (so in other words, just two values, each being raised to some power) and you're trying to find

$$\sum_{k=1}^n ^nC_k.3^k$$ to create a superscript in $\LaTeX$ just use ^ (and add {} for multiple characters) before it

Notice that we can see a=3, but b isn't present. In fact, it's just hidden as b=1 because

$$\sum_{k=1}^n ^nC_k.3^k1^{n-k}$$

is exactly the same thing. So our final answer would be

$$\sum_{k=0}^n ^nC_k.3^k-^nC_03^0=(3+1)^n-1=4^n-1$$

EDIT: and it seems that latex has changed, once again... Man I'm getting annoyed with it. Give me a second and I'll figure it out.
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P: 11,866
 Quote by Pranav-Arora Would you please tell me how to make the "n" before "C" in Superscript?
You can use {^n}C_k or {}^n C_k.
P: 3,807
 Quote by Mentallic Notice that we can see a=3, but b isn't present. In fact, it's just hidden as b=1 because $$\sum_{k=1}^n ^nC_k.3^k1^{n-k}$$ is exactly the same thing. So our final answer would be $$\sum_{k=0}^n ^nC_k.3^k-^nC_03^0=(3+1)^n-1=4^n-1$$ EDIT: and it seems that latex has changed, once again... Man I'm getting annoyed with it. Give me a second and I'll figure it out.
Thanks Mentallic but i don't understand from where you got
$$-^nC_03^0$$

 Quote by vela You can use {^n}C_k or {}^n C_k.
Thanks vela, it worked
 HW Helper P: 6,189 Hi Pranav-Arora! Seeing that Mentallic and vela are not around, I'll answer your question. It's part of the sigma notation and its implications. In particular this is about the boundaries of the sum, which in your case is starting with k=1. What you have is: $$\begin{eqnarray} \sum_{k=1}^n {^n}C_k \cdot 3^k &=&{^n}C_1 \cdot 3^1 + {^n}C_2 \cdot 3^2 + \dotsb + {^n}C_n \cdot 3^n\\ &=&({^n}C_0 \cdot 3^0 + {^n}C_1 \cdot 3^1 + {^n}C_2 \cdot 3^2 + \dotsb + {^n}C_n \cdot 3^n) - {^n}C_0 \cdot 3^0 \\ &=&(\sum_{k=0}^n {^n}C_k \cdot 3^k) - {^n}C_0 \cdot 3^0 \end{eqnarray}$$ The way to change the boundaries is always the same. Your write out the sum in its terms, change what you want to change, and change it back again into sigma notation. Note that the sigma notation is only a shorthand notation. Don't think of it as something magical that has its own rules - it hasn't. It's just shorthand.
P: 3,807
 Quote by I like Serena Hi Pranav-Arora! Seeing that Mentallic and vela are not around, I'll answer your question. It's part of the sigma notation and its implications. In particular this is about the boundaries of the sum, which in your case is starting with k=1. What you have is: $$\begin{eqnarray} \sum_{k=1}^n {^n}C_k \cdot 3^k &=&{^n}C_1 \cdot 3^1 + {^n}C_2 \cdot 3^2 + ... + {^n}C_n \cdot 3^n\\ &=&({^n}C_0 \cdot 3^0 + {^n}C_1 \cdot 3^1 + {^n}C_2 \cdot 3^2 + ... + {^n}C_n \cdot 3^n) - {^n}C_0 \cdot 3^0 \\ &=&(\sum_{k=0}^n {^n}C_k \cdot 3^k) - {^n}C_0 \cdot 3^0 \end{eqnarray}$$ The way to change the boundaries is always the same. Your write out the sum in its terms, change what you want to change, and change it back again into sigma notation. Note that the sigma notation is only a shorthand notation. Don't think of it as something magical that has its own rules - it hasn't. It's just shorthand.
But what i would do if a question appears like this:-

$$\sum^n_{k=0} (2k+1) {}^nC_k$$.
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P: 6,189
 Quote by Pranav-Arora Thanks for your reply I like Serena. But what i would do if a question appears like this:- $$\sum^n_{k=0} (2k+1) {}^nC_k$$.
Ah, this one is a bit more difficult.
The method I know is to define a function of x and integrate it.
That is:
$$s(x) = \sum^n_{k=0} {}^nC_k (2k+1) x^{2k}$$
The result you're looking for in this case is s(1).

If you integrate it, you should find a form that looks more like your previous problem.
You can rewrite that without the sigma and binomium.
Afterward you differentiate again.
And finally you fill in the value 1.

Care to try?
P: 3,807
 Quote by I like Serena Ah, this one is a bit more difficult. The method I know is to define a function of x and integrate it. That is: $$s(x) = \sum^n_{k=0} {}^nC_k (2k+1) x^{2k}$$ The result you're looking for in this case is s(1). If you integrate it, you should find a form that looks more like your previous problem. You can rewrite that without the sigma and binomium. Afterward you differentiate again. And finally you fill in the value 1. Care to try?
How you get x2k?
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P: 3,562
 Quote by I like Serena Seeing that Mentallic and vela are not around, I'll answer your question.
Good thing you did considering I was miles away from any internet connections (locked away in my room, sleeping)

 Quote by Pranav-Arora How you get x2k?
What happens if you integrate x2k?
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P: 11,866
 Quote by Pranav-Arora How you get x2k?
Is this problem from a pre-calculus class since you posted in the precalc forum? If so, did you learn any identities involving binomial coefficients?
P: 3,807
 Quote by Mentallic Good thing you did considering I was miles away from any internet connections (locked away in my room, sleeping) What happens if you integrate x2k?
Maybe
$$\frac{2(x^{2k+1})}{2k+1}$$
P: 3,807
 Quote by vela Is this problem from a pre-calculus class since you posted in the precalc forum? If so, did you learn any identities involving binomial coefficients?
Yep, i have learned some identities involving binomial coefficients but i didn't knew that to solve this question we have to perform integration and differetiation.
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 Quote by Pranav-Arora Maybe $$\frac{2(x^{2k+1})}{2k+1}$$
Not quite. Take the derivative of that to see where you went wrong.

 Quote by Pranav-Arora Yep, i have learned some identities involving binomial coefficients but i didn't knew that to solve this question we have to perform integration and differetiation.
They might have given you a formula to use?
P: 3,807
 Quote by Mentallic Not quite. Take the derivative of that to see where you went wrong.
I took the derivative i again got x2k. What is your answer when you integrate x2k?

 Quote by Mentallic They might have given you a formula to use?
I have only one formula which involves integration within limits. But i only know how to integrate without limits. I have never solved questions which involve integration within limits.
Here's the formula:-

$$(1+x)^n=C_0+C_1x+C_2x^2+........+C_kx^k+.......+C_nx^n...$$

Integrating the above equation with respect to x between limits 0 to 1, (I don't understand what it is ) we get,

$$(\frac {(1+x)^{n+1}}{n+1})_0^1=(C_0x+C_1\frac{x^2}{2}+C_2\frac{x^3}{3}+.....C_ n\frac{x^{n+1}}{n+1})_0^1$$

$$=C_0+\frac{C_1}{2}+\frac{C_2}{3}.....\frac{C_n}{n+1}=\frac{2^{2n+1}-1}{n+1}$$
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P: 3,562
 Quote by Pranav-Arora I took the derivative i again got x2k. What is your answer when you integrate x2k?
The integral of xn is $$\frac{x^{n+1}}{n+1}$$ where n is a constant. Since k is just some constant, the same rule applies. It looks as though you're treating k as if it's a variable.

I have only one formula which involves integration within limits. But i only know how to integrate without limits. I have never solved questions which involve integration within limits.
Here's the formula:-

$$(1+x)^n=C_0+C_1x+C_2x^2+........+C_kx^k+.......+C_nx^n...$$[/quote]

If that is the formula, then when evaluating this at 1, we would have

$$(1+1)^n=2^n=C_0+C_1+C_2+...+C_k+...+C_n$$ correct?

and evaluating it at 0

$$(1+0)^n=1^n=1=C_0+C_1\cdot 0+C_2\cdot 0+...=1$$

 Quote by Pranav-Arora Integrating the above equation with respect to x between limits 0 to 1, (I don't understand what it is )
What don't you understand about it? If two sides are equal, then the integral of both sides will be equal (disregarding the constant of integration).

 Quote by Pranav-Arora $$=C_0+\frac{C_1}{2}+\frac{C_2}{3}.....\frac{C_n}{n+1}=\frac{2^{2n+1}-1}{n+1}$$
So can you apply this to your question now?
P: 3,807
 Quote by Mentallic So can you apply this to your question now?
How would i apply this to my question?

I like Serena said that define a function of x and integrate it but i still don't get how he got x2k?

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