# Homework Help: Converting Sigma notation

1. Jun 28, 2011

### Saitama

Converting Sigma notation....

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

Last edited: Jun 28, 2011
2. Jun 28, 2011

### Mentallic

Re: Converting Sigma notation....

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...

I'm still not quite getting it. That is just an expression, but what is the actual question?

3. Jun 28, 2011

### Saitama

Re: Converting Sigma notation....

Sorry!! I incorrectly wrote down the binomial expansion.

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.

4. Jun 28, 2011

### Mentallic

Re: Converting Sigma notation....

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.

5. Jun 28, 2011

### vela

Staff Emeritus
Re: Converting Sigma notation....

You can use {^n}C_k or {}^n C_k.

6. Jun 28, 2011

### Saitama

Re: Converting Sigma notation....

Thanks Mentallic but i don't understand from where you got
$$-^nC_03^0$$

Thanks vela, it worked

7. Jun 28, 2011

### I like Serena

Re: Converting Sigma notation....

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.

8. Jun 28, 2011

### Saitama

Re: Converting Sigma notation....

But what i would do if a question appears like this:-

$$\sum^n_{k=0} (2k+1) {}^nC_k$$.

9. Jun 28, 2011

### I like Serena

Re: Converting Sigma notation....

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?

10. Jun 28, 2011

### Saitama

Re: Converting Sigma notation....

How you get x2k?

11. Jun 28, 2011

### Mentallic

Re: Converting Sigma notation....

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?

12. Jun 28, 2011

### vela

Staff Emeritus
Re: Converting Sigma notation....

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?

13. Jun 28, 2011

### Saitama

Re: Converting Sigma notation....

Maybe
$$\frac{2(x^{2k+1})}{2k+1}$$

14. Jun 28, 2011

### Saitama

Re: Converting Sigma notation....

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.

15. Jun 28, 2011

### Mentallic

Re: Converting Sigma notation....

Not quite. Take the derivative of that to see where you went wrong.

They might have given you a formula to use?

16. Jun 29, 2011

### Saitama

Re: Converting Sigma notation....

I took the derivative i again got x2k. What is your answer when you integrate x2k?

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,

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

$$=C_0+\frac{C_1}{2}+\frac{C_2}{3}.....\frac{C_n}{n+1}=\frac{2^{2n+1}-1}{n+1}$$

17. Jun 29, 2011

### Mentallic

Re: Converting Sigma notation....

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

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

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).

So can you apply this to your question now?

18. Jun 29, 2011

### Saitama

Re: Converting Sigma notation....

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?

19. Jun 29, 2011

### I like Serena

Re: Converting Sigma notation....

I defined an arbitrary function s(x) that looks a bit like your problem with the special property that if you substitute x=1, it is identical to your problem.

It's a trick to solve your problem.

The choice of the power 2k was inspired so that the factor (2k+1) would disappear during integration.

20. Jun 29, 2011

### Saitama

Re: Converting Sigma notation....

Ok i got it!!
But how i would integrate the expression. It involves sigma notation and i have never done integration of any expression which involves sigma notation.

21. Jun 29, 2011

### I like Serena

Re: Converting Sigma notation....

So write out the terms of the summation, do the integration, and combine the resulting terms back into sigma notation.

22. Jun 29, 2011

### Saitama

Re: Converting Sigma notation....

I did as you said. After integrating, i got
$$x+nx^3+\frac{n(n-1)x^5}{2!}+\frac{n(n-1)(n-2)x^7}{3!}........x^{2n+1}$$

Now what should i do next?

23. Jun 29, 2011

### Mentallic

Re: Converting Sigma notation....

To be frank, I have never answered such questions myself either, but I took I like Serena's advice and it worked out wonderfully.

We'll start again just to make things clear,

To solve $$\sum_{k=0}^{n}(2k+1)^{n}C_k$$

we define a function

$$f(x)=\sum_{k=0}^{n}(2k+1){^n}C_kx^{2k}$$

we make it x2k because the integral of that is $\frac{x^{2k+1}}{2k+1}$ and notice how that denominator will cancel with the (2k+1) factor in the original question. So we have,

$$\int{f(x)dx}=\sum_{k=0}^{n}{^n}C_kx^{2k+1}$$

And here is the tricky part, we need to convert the right side into a binomial expression using the formula

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

It is clear that the b is again missing, which it is just hidden as b=1, but we need to convert the x2k+1 in such a way that it is equivalent to ak.

Use your rules for indices to convert it in such a way.

Last edited: Jun 29, 2011
24. Jun 29, 2011

### Mentallic

Re: Converting Sigma notation....

What might have been easier for you is you could've split the summation into

$$2\sum k{^n}C_k+\sum {^n}C_k$$

and then defined $$s(x)=2\sum k{^n}C_kx^{k+1}+\sum {^n}C_kx^{k+1}$$

25. Jun 29, 2011

### I like Serena

Re: Converting Sigma notation....

$$\sum_{k=0}^n {^n}C_k (2k+1) x^{2k} = {^n}C_0 + {^n}C_1 \cdot 3 x^2 + {^n}C_2 \cdot 5 x^4 + ... + {^n}C_k (2k + 1) x^{2k} + ...$$
$${^n}C_0 \cdot x + {^n}C_1 \cdot x^3 + {^n}C_2 \cdot x^5 + ... + {^n}C_k x^{2k+1} + ...$$
$$\sum_{k=0}^n {^n}C_k x^{2k+1}$$