Converting Sigma notation....

Pranav-Arora

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 x^2k?

I like Serena

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.

Pranav-Arora

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.

I like Serena

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

Pranav-Arora

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

Now what should i do next?

Mentallic

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 [tex]\sum_{k=0}^{n}(2k+1)^{n}C_k[/tex]

we define a function

[tex]f(x)=\sum_{k=0}^{n}(2k+1){^n}C_kx^{2k}[/tex]

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

[tex]\int{f(x)dx}=\sum_{k=0}^{n}{^n}C_kx^{2k+1}[/tex]

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

[tex]\sum_{k=0}^{n}{^n}C_ka^kb^{n-k}=(a+b)^n[/tex]

It is clear that the b is again missing, which it is just hidden as b=1, but we need to convert the x^2k+1 in such a way that it is equivalent to a^k.

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

Mentallic

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

[tex]2\sum k{^n}C_k+\sum {^n}C_k[/tex]

and then defined [tex]s(x)=2\sum k{^n}C_kx^{k+1}+\sum {^n}C_kx^{k+1}[/tex]

I like Serena

Hmm, what I intended was this:

[tex]\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} + ... [/tex]

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

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

Pranav-Arora

Would it be like this (x^2k+1+1)ⁿ.

Pranav-Arora

I did the same way.

Mentallic

Noo...

[tex]\sum{^n}C_kx^k1^{n-k}=(x+1)^n[/tex]

and not [tex](x^k+1)^n[/tex]

Use the fact that

[tex]a^{b+1}=a\cdot a^b[/tex] and [tex]a^{2b}=\left(a^2\right)^b[/tex]

Pranav-Arora

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

[tex]x^{2k+1}=x^{2k}.x=(x^2)^k.x[/tex]

Mentallic

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

So now we have

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

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

[tex]x\sum_{k=0}^{n}{^n}C_k\left(x^2\right)^k1^{n-k}[/tex]

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).

Pranav-Arora

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

I substitute the value 1 and i get
[tex]2^n[/tex]

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

Good!

Yep!

Mentallic

Yep, that's what I meant by



You're nearly there!

Pranav-Arora

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

Now i substituted the value 1 and i got:-
[tex]2n.2^{n-1}[/tex]

Right...?