# Converting Sigma Notation: Simplifying Binomial Theorem

• Saitama
What would you do in this case?Thanks for your reply I like Serena. :smile:But what i would do if a question appears like this:-\sum^n_{k=0} (2k+1) {}^nC_k....and then the options were in terms of 2^n or 2^n+1 or 2^n+2 or 2^n+3What would you do in this case?I think the key to this question is to remember that the binomial theorem is:\sum_{k=0}^n {^n
Saitama
Converting Sigma notation...

## Homework Statement

(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?

## The Attempt at a Solution

Last edited:

Pranav-Arora said:

## Homework Statement

(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...

Pranav-Arora said:
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?

Mentallic said:
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.

Mentallic said:
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.

Pranav-Arora said:
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.

Pranav-Arora said:
Would you please tell me how to make the "n" before "C" in Superscript?
You can use {^n}C_k or {}^n C_k.

Mentallic said:
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
$$-^nC_03^0$$

vela said:
You can use {^n}C_k or {}^n C_k.
Thanks vela, it worked

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.

I like Serena said:
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.

Thanks for your reply I like Serena.
But what i would do if a question appears like this:-

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

Pranav-Arora said:
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?

I like Serena said:
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?

I like Serena said:
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)

Pranav-Arora said:
How you get x2k?

What happens if you integrate x2k?

Pranav-Arora said:
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?

Mentallic said:
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
$$\frac{2(x^{2k+1})}{2k+1}$$

vela said:
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.

Pranav-Arora said:
Maybe
$$\frac{2(x^{2k+1})}{2k+1}$$
Not quite. Take the derivative of that to see where you went wrong.

Pranav-Arora said:
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?

Mentallic said:
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?

Mentallic said:
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,

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

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

Pranav-Arora said:
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

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

Pranav-Arora said:
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).

Pranav-Arora said:
$$=C_0+\frac{C_1}{2}+\frac{C_2}{3}...\frac{C_n}{n+1}=\frac{2^{2n+1}-1}{n+1}$$
So can you apply this to your question now?

Mentallic said:
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?

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

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.

I like Serena said:
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.

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.

Pranav-Arora said:
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.

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

I like Serena said:
So write out the terms of the summation, do the integration, and combine the resulting terms back into 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?

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:

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

Pranav-Arora said:
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?

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

Integration would give:
$${^n}C_0 \cdot x + {^n}C_1 \cdot x^3 + {^n}C_2 \cdot x^5 + ... + {^n}C_k x^{2k+1} + ...$$

Converting back to sigma notation:
$$\sum_{k=0}^n {^n}C_k x^{2k+1}$$

Mentallic said:
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.

Would it be like this (x2k+1+1)n.

I like Serena said:

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

Integration would give:
$${^n}C_0 \cdot x + {^n}C_1 \cdot x^3 + {^n}C_2 \cdot x^5 + ... + {^n}C_k x^{2k+1} + ...$$

Converting back to sigma notation:
$$\sum_{k=0}^n {^n}C_k x^{2k+1}$$

I did the same way.

Pranav-Arora said:
Would it be like this (x2k+1+1)n.

Noo...

$$\sum{^n}C_kx^k1^{n-k}=(x+1)^n$$

and not $$(x^k+1)^n$$

Use the fact that

$$a^{b+1}=a\cdot a^b$$ and $$a^{2b}=\left(a^2\right)^b$$

Mentallic said:
Noo...

$$\sum{^n}C_kx^k1^{n-k}=(x+1)^n$$

and not $$(x^k+1)^n$$

Use the fact that

$$a^{b+1}=a\cdot a^b$$ and $$a^{2b}=\left(a^2\right)^b$$

I tried it but got stuck again. I did it like this:-

$$x^{2k+1}=x^{2k}.x=(x^2)^k.x$$

Pranav-Arora said:
I tried it but got stuck again. I did it like this:-

$$x^{2k+1}=x^{2k}.x=(x^2)^k.x$$

Why did you get stuck? That's exactly what it should be!

So now we have

$$\int{f(x)dx}=\sum_{k=0}^{n}{^n}C_kx\cdot \left(x^2\right)^k$$

And since x is independent of k, it can move out the front of the summation, so we have

$$x\sum_{k=0}^{n}{^n}C_k\left(x^2\right)^k1^{n-k}$$

And now apply the formula to convert it into a binomial. And since you need to find f(1), take the derivative of both sides to get the expression for f(x).

Mentallic said:
Why did you get stuck? That's exactly what it should be!

So now we have

$$\int{f(x)dx}=\sum_{k=0}^{n}{^n}C_kx\cdot \left(x^2\right)^k$$

And since x is independent of k, it can move out the front of the summation, so we have

$$x\sum_{k=0}^{n}{^n}C_k\left(x^2\right)^k1^{n-k}$$

And now apply the formula to convert it into a binomial. And since you need to find f(1), take the derivative of both sides to get the expression for f(x).

If i convert it into binomial, i get
$$x.(x^2+1)^n$$

I substitute the value 1 and i get
$$2^n$$

But then how i would find out the derivative??
Do i have to first take the derivative and substitute the value 1?

Pranav-Arora said:
If i convert it into binomial, i get
$$x.(x^2+1)^n$$

Good!

Pranav-Arora said:
But then how i would find out the derivative??
Do i have to first take the derivative and substitute the value 1?

Yep!

Pranav-Arora said:
But then how i would find out the derivative??
Do i have to first take the derivative and substitute the value 1?
Yep, that's what I meant by

Mentallic said:
And since you need to find f(1), take the derivative of both sides to get the expression for f(x).

You're nearly there!

Thanks!
I think that this time i am right.
I took the derivative and found it to be
$$2nx(x^2+1)^{n-1}$$

Now i substituted the value 1 and i got:-
$$2n.2^{n-1}$$

Right...?

No, not quite.
What you have is not the derivative of $x.(x^2+1)^n$.

You need to apply the so called product rule.
That is: (u v)' = u' v + u v'
And you have to apply the so called chain rule.
That is: (u(v))' = u'(v) v'

Are you familiar with those rules?

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