Number Theory: Show that (nC0) -(1/2)(nC1)+(1/3)(nC2)-....+((-1)n/n)(nCn) = 1/(n+1)

Steven G

Steven G

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Can I please get a hint on this one?
Show that (nC0) -(1/2)(nC1)+(1/3)(nC2)-....+((-1)n/n)(nCn) = 1/(n+1)
 
Steven G said:
Show that (nC0) -(1/2)(nC1)+(1/3)(nC2)-....+((-1)n/n)(nCn) = 1/(n+1)
Click to expand...
The last term does look consistent with previous ones, e.g. (1/3)(nC2) vs. (1/n)(nCn) -- shouldn't the last one be (1/(n+1))(nCn) ?
 
Assuming you mean [imath]\sum_{k=0}^n (-1)^k{k \choose n}\frac{1}{k+1}[/imath] look at [imath]\frac{n+1}{k+1} {k \choose n}[/imath]
 
I hope this is not your home work because I posting a solution :):
[math]\sum_{k=0}^n (-1)^k{k \choose n} \frac{1}{k+1} = \sum_{k=0}^n (-1)^k\frac{n!}{k!(n-k)!} \frac{1}{k+1}[/math][math]= \sum_{k=0}^n (-1)^k\frac{n!}{(k+1)!(n-k)!} = \frac{1}{n+1} \sum_{k=0}^n (-1)^k{k+1 \choose n+1}[/math][math]= \frac{1}{n+1} \left( \sum_{k=0}^n (-1)^k{k+1 \choose n+1} \right)[/math]-- replacing [imath]k+1[/imath] with [imath]j[/imath]:
[math]= \frac{1}{n+1} \left( \sum_{j=1}^{n+1} (-1)^{j-1}{j \choose n+1} \right) = -\frac{1}{n+1} \left( \sum_{j=1}^{n+1} (-1)^{j}{j \choose n+1} \right)[/math][math]= -\frac{1}{n+1} \left( -1 + \sum_{j=0}^{n+1} (-1)^{j}{j \choose n+1} \right) = -\frac{1}{n+1} \left( -1 + \left( 1 - 1\right)^{n+1} \right) = \frac{1}{n+1}[/math]
 
blamocur said:
I hope this is not your home work because I posting a solution :):
[math]\sum_{k=0}^n (-1)^k{k \choose n} \frac{1}{k+1} = \sum_{k=0}^n (-1)^k\frac{n!}{k!(n-k)!} \frac{1}{k+1}[/math][math]= \sum_{k=0}^n (-1)^k\frac{n!}{(k+1)!(n-k)!} = \frac{1}{n+1} \sum_{k=0}^n (-1)^k{k+1 \choose n+1}[/math][math]= \frac{1}{n+1} \left( \sum_{k=0}^n (-1)^k{k+1 \choose n+1} \right)[/math]-- replacing [imath]k+1[/imath] with [imath]j[/imath]:
[math]= \frac{1}{n+1} \left( \sum_{j=1}^{n+1} (-1)^{j-1}{j \choose n+1} \right) = -\frac{1}{n+1} \left( \sum_{j=1}^{n+1} (-1)^{j}{j \choose n+1} \right)[/math][math]= -\frac{1}{n+1} \left( -1 + \sum_{j=0}^{n+1} (-1)^{j}{j \choose n+1} \right) = -\frac{1}{n+1} \left( -1 + \left( 1 - 1\right)^{n+1} \right) = \frac{1}{n+1}[/math]
Click to expand...
I will see if I now can do this on my own. There are just too maybe typos to fix your work! Although, it is an excellent hint!
Thanks!
 
blamocur said:
I hope this is not your home work because I posting a solution :):
[math]\sum_{k=0}^n (-1)^k{k \choose n} \frac{1}{k+1} = \sum_{k=0}^n (-1)^k\frac{n!}{k!(n-k)!} \frac{1}{k+1}[/math][math]= \sum_{k=0}^n (-1)^k\frac{n!}{(k+1)!(n-k)!} = \frac{1}{n+1} \sum_{k=0}^n (-1)^k{k+1 \choose n+1}[/math][math]= \frac{1}{n+1} \left( \sum_{k=0}^n (-1)^k{k+1 \choose n+1} \right)[/math]-- replacing [imath]k+1[/imath] with [imath]j[/imath]:
[math]= \frac{1}{n+1} \left( \sum_{j=1}^{n+1} (-1)^{j-1}{j \choose n+1} \right) = -\frac{1}{n+1} \left( \sum_{j=1}^{n+1} (-1)^{j}{j \choose n+1} \right)[/math][math]= -\frac{1}{n+1} \left( -1 + \sum_{j=0}^{n+1} (-1)^{j}{j \choose n+1} \right) = -\frac{1}{n+1} \left( -1 + \left( 1 - 1\right)^{n+1} \right) = \frac{1}{n+1}[/math]
Click to expand...
It's my homework but no one is going to grade it.
 
I don't see why I couldn't see this on my own. I saw many others! Thanks for your help!
 
Steven G said:
Can I please get a hint on this one?
Show that (nC0) -(1/2)(nC1)+(1/3)(nC2)-....+((-1)n/n)(nCn) = 1/(n+1)
Click to expand...
What are the unknowns (variables) of this problem?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this problem
 
khansaheb said:
What are the unknowns (variables) of this problem?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this problem
Click to expand...
I already got the answer so I don't need to follow your rules!o_O
 
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