
#19
Jun2911, 01:37 AM

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P: 6,189

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
Jun2911, 01:44 AM

P: 3,545

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
Jun2911, 01:57 AM

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P: 6,189





#22
Jun2911, 02:08 AM

P: 3,545

[tex]x+nx^3+\frac{n(n1)x^5}{2!}+\frac{n(n1)(n2)x^7}{3!}........x^{2n+1}[/tex] Now what should i do next? 



#23
Jun2911, 02:15 AM

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P: 3,433

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^{nk}=(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. 



#24
Jun2911, 02:20 AM

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P: 3,433

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] 



#25
Jun2911, 02:24 AM

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P: 6,189

[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] 



#28
Jun2911, 03:07 AM

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P: 3,433

[tex]\sum{^n}C_kx^k1^{nk}=(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] 



#29
Jun2911, 03:39 AM

P: 3,545

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



#30
Jun2911, 04:54 AM

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P: 3,433

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^{nk}[/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). 



#31
Jun2911, 05:07 AM

P: 3,545

[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? 



#33
Jun2911, 07:52 AM

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P: 3,433

You're nearly there! 



#34
Jun2911, 09:39 AM

P: 3,545

Thanks!!
I think that this time i am right. I took the derivative and found it to be [tex]2nx(x^2+1)^{n1}[/tex] Now i substituted the value 1 and i got: [tex]2n.2^{n1}[/tex] Right...? 



#35
Jun2911, 10:59 AM

HW Helper
P: 6,189

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