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MC III Combinatorics help

 
 
SongKevin的头像
MC III Combinatorics help
SongKevin - 2023年01月15日 Sunday 09:46
 

I need help with 3.17 and 3.20, thanks

 
ProfessorAreteem的头像
Re: MC III Combinatorics help
ProfessorAreteem - 2023年01月17日 Tuesday 13:04
 

Here's some hints in case they're still needed.

For 3.17: Let y = sqrt(x) and try writing out the binomial theorem for (y+2)^(2n+1). Note if x has an integer power, then y is raised to an even power. What happens if we let y = 1 and y = -1 in the binomial theorem?

For 3.20: For this one as well, write out the binomial theorem. Note that each term can be simplified to +- (10 choose k) * (a power of 2) * (a power of x) for some k = 0 to 10. Since (10 choose k) is always an integer, consider when (a power of 2) is rational.

SongKevin的头像
Re: MC III Combinatorics help
SongKevin - 2023年01月20日 Friday 20:56
 

is it 

3.17: 

\( (3^{2n+1}+1)/2 \)


3.20:

\( 3360x^6 \), 1/32

?

ProfessorAreteem的头像
Re: MC III Combinatorics help
ProfessorAreteem - 2023年01月23日 Monday 10:29
 

3.17 looks good.

For 3.20, 1/32 is correct, as is the fact we have x^6 term. Where did your 3360 come from? The coefficient there should actually be 210.