Converting Sigma Notation: Simplifying Binomial Theorem

[Saitama]

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?

 

[I like Serena]

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

Good!

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

Yep!

 

[Mentallic]

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

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!

 

[Saitama]

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...?

 

[I like Serena]

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?

 

[Saitama]

I am not familiar with the product rule but when i calculated the derivative on Wolfram, it was the same as i got?

 

[I like Serena]

I am not familiar with the product rule but when i calculated the derivative on Wolfram, it was the same as i got?

I get a different result from wolfram:
http://www.wolframalpha.com/input/?i=d%2Fdx+%28x+%28x^2%2B1%29^n%29

 

[Mentallic]

I am not familiar with the product rule

Then you need to either go back and run over it again or take our word for it

Try applying the product rule, let u=x and v=(x^2+1)^n

 

[Saitama]

I again tried to calculate the derivative and got:-
2nx^2(x^2+1)^{n-1}+(x^2+1)^n

Right...?
Now i substitute the value 1 and i get:-
2^n(1+n)

Am i right this time?

 

[I like Serena]

Right... and we're done!

 

[Saitama]

Right... and we're done!

Thanks for reply!
I have more questions.
The question is:-

\sum_{k=1}^n \frac{{}^nC_k}{k+2}

Can you provide me some hints to start?
(Please don't think that i am trying to solve my homework, it's not a homework question, it is from a test paper, i am only trying to get a strong hold on binomial theorem and Calculus )

 

[I like Serena]

Looks like a combination of the previous 2 problems.

The boundary k=1 is one off. You need to correct that.

And you can define a function of x, such that the denominator disappears when you take the derivative.

 

[Saitama]

Looks like a combination of the previous 2 problems.

The boundary k=1 is one off. You need to correct that.

And you can define a function of x, such that the denominator disappears when you take the derivative.

I thought of that but how should i define a function of x? Like in previous question you took x^2k, so what should i take here?

 

[Mentallic]

I thought of that but how should i define a function of x? Like in previous question you took x^2k, so what should i take here?

Last time we chose x^2k because the integral of it caused us to cancel the factor 2k+1. This time we will be taking the derivative and what n must be chosen such that when we take the derivative of xⁿ, it cancels out the factor \frac{1}{k+2}

And by the way, you are always allowed to check your answer, so whatever you choose, take the derivative to see what happens

 

[Saitama]

Last time we chose x^2k because the integral of it caused us to cancel the factor 2k+1. This time we will be taking the derivative and what n must be chosen such that when we take the derivative of xⁿ, it cancels out the factor \frac{1}{k+2}

And by the way, you are always allowed to check your answer, so whatever you choose, take the derivative to see what happens

I tried solving it, please tell me if i am wrong somewhere:-

Define it as a function of x and differentiate it, i.e.
f(x)=\sum_{k=1}^n \frac{{}^nC_k}{k+2}x^{k+2}

I mould it so that it is in the form where k starts from 0, i.e
f(x)=\sum_{k=0}^n \frac{{}^nc_k}{k+2}-\frac{{}^nc_0x^2}{2}

Then i differentiated it and i get:-
\frac{d}{dx}(f(x))=(\sum_{k=0}^n{}^nC_kx^{k+1})-{}^nC_0x

Converting it to binomial form:-
\frac{d}{dx}(f(x))=x(x+1)^n-x

Now what should be the next step?
Should the next step be integration?

 

[Mentallic]

I tried solving it, please tell me if i am wrong somewhere:-

Define it as a function of x and differentiate it, i.e.
f(x)=\sum_{k=1}^n \frac{{}^nC_k}{k+2}x^{k+2}

I mould it so that it is in the form where k starts from 0, i.e
f(x)=\sum_{k=0}^n \frac{{}^nc_k}{k+2}x^{k+2}-\frac{{}^nc_0x^2}{2}

Then i differentiated it and i get:-
\frac{d}{dx}(f(x))=(\sum_{k=0}^n{}^nC_kx^{k+1})-{}^nC_0x

Converting it to binomial form:-
\frac{d}{dx}(f(x))=x(x+1)^n-x

Now what should be the next step?
Should the next step be integration?

Beautifully done so far. You just accidentally forgot to write down the red part.
Well yes, of course it's time for integration since your ultimate goal is to calculate f(1)

 

[Saitama]

Well yes, of course it's time for integration since your ultimate goal is to calculate f(1)

Thanks Mentallic and I Like Serena!
I think i have got it right.
Integrating x(x+1)^n-x, i get:-

\frac{x(x+1)^{n+1}}{n+1}-\frac{(x+1)^{n+2}}{(n+1)(n+2)}

Substituting the value 1, i get:-
\frac{n.2^{n+1}}{(n+1)(n+2)}

Am i right...?

Beautifully done so far. You just accidentally forgot to write down the red part.

Oops! Sorry..

 

[I like Serena]

Almost. ;)

First, it seems that you forgot to integrate the loose term "x".

And when you integrate, there is a slight problem called the integration constant.
That is, you need to add the as yet unknown constant C to the result of the integration.

Then you need to fill in a value for n and see how it works out.
From the result you can determine the value of C.

 

[Saitama]

Almost. ;)

First, it seems that you forgot to integrate the loose term "x".

And when you integrate, there is a slight problem called the integration constant.
That is, you need to add the as yet unknown constant C to the result of the integration.

Then you need to fill in a value for n and see how it works out.
From the result you can determine the value of C.

I didn't get how to find out the constant?
Where i forgot to integrate "x"?

 

[I like Serena]

Where i forgot to integrate "x"?

You had to subtract the k=0 term, which ended up being just "x".
I don't see it back in your integrated result.

I didn't get how to find out the constant?

Let me explain with an example.

Suppose you have the function f(x)=x² + 3
Now you know that f(0)=3 don't you?

Taking the derivative we get f'(x)=2x.
Taking the integral again we get \int f'(x)dx = x² + C
(This is called an indefinite integral, since the boundaries were not given.)

As you can see the result is not equal to the original function - we lost the '3' in the process.
But since we know that f(0)=3, we can deduce that the integration constant C must be 3.

 

[Saitama]

You had to subtract the k=0 term, which ended up being just "x".
I don't see it back in your integrated result.




Let me explain with an example.

Suppose you have the function f(x)=x² + 3
Now you know that f(0)=3 don't you?

Taking the derivative we get f'(x)=2x.
Taking the integral again we get \int f'(x)dx = x² + C
(This is called an indefinite integral, since the boundaries were not given.)

As you can see the result is not equal to the original function - we lost the '3' in the process.
But since we know that f(0)=3, we can deduce that the integration constant C must be 3.

In which step should i substitute zero to get the constant?

 

[I like Serena]

In which step should i substitute zero to get the constant?

I'm assuming you mean in my example.
(It's slightly different for the current problem.)

In my example we have as input f(x) = \int f'(x) dx = x^2 + C and f(0)=3.

We want to find f(x).

Filling in x=0 gives: f(0) = 0^2 + C = 3.

So C = 3.

 

[Saitama]

I'm assuming you mean in my example.
(It's slightly different for the current problem.)

In my example we have as input f(x) = \int f'(x) dx = x^2 + C and f(0)=3.

We want to find f(x).

Filling in x=0 gives: f(0) = 0^2 + C = 3.

So C = 3.

I got what you said.
But i am asking in which step should i put x=0?
Should i put x=0 during the integration step?

 

[I like Serena]

I got what you said.
But i am asking in which step should i put x=0?
Should i put x=0 during the integration step?

I don't understand your question.
Afaik I have explicitly stated where you put x=0.

I only could have added as a last step that the conclusion is that f(x)=x² + 3.

 

[Saitama]

i think the constant is zero.

 

[I like Serena]

i think the constant is zero.

Now I really don't understand you.

Which constant?
In which formula?
In which post?
Why would it be zero?

 

[Saitama]

Now I really don't understand you.

Which constant?
In which formula?
In which post?
Why would it be zero?

You gave an example that f(x)=x²+3.
And you said that f(0)=3, i.e you substituted the value 0 at the place of x. Right..?
Now that 3 is our integration constant, we can substitute 3 in \int f'(x)=x^2+C at the place of C.

I did the same in my question. I substituted 0 in f(x), i.e i substituted 0 in:-
f(x)=\sum_{k=1}^n \frac{{}^nC_k}{k+2}x^{k+2}

So i got f(0)=0. Therefore integration constant is 0.

Did you get me now?

 

[I like Serena]

Yes, I more or less get what you did, but it is not right.Let's take it a couple of steps back.

You had:
f'(x) = x(x+1)^n \color{red}{\textbf{- x}}

You integrated this wonderfully, but you forgot the "\color{red}{\textbf{- x}}" I just marked.

So you should have:
f(x) = \int f'(x)dx =\frac{x(x+1)^{n+1}}{n+1}-\frac{(x+1)^{n+2}}{(n+1)(n+2)} - \frac {x^2} 2 + C

Now you need to determine the integration constant C using f(0) = 0, but C is not zero!

 

[Saitama]

Yes, I more or less get what you did, but it is not right.


Let's take it a couple of steps back.

You had:
f'(x) = x(x+1)^n \color{red}{\textbf{- x}}

You integrated this wonderfully, but you forgot the "\color{red}{\textbf{- x}}" I just marked.

So you should have:
f(x) = \int f'(x)dx =\frac{x(x+1)^{n+1}}{n+1}-\frac{(x+1)^{n+2}}{(n+1)(n+2)} - \frac {x^2} 2 + C

Now you need to determine the integration constant C using f(0) = 0, but C is not zero!

So then what's the C, i don't seem to find any other way to find C.

 

[I like Serena]

So then what's the C, i don't seem to find any other way to find C.

What do you get if you fill in x=0 in this formula?
f(x) = \frac{x(x+1)^{n+1}}{n+1}-\frac{(x+1)^{n+2}}{(n+1)(n+2)} - \frac {x^2} 2 + C