anonymous
  • anonymous
someone could help me here , please .
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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anonymous
  • anonymous
i don't know what method i have to use there. :l
UsukiDoll
  • UsukiDoll
I only have a slight idea for the first one given n is an odd integer \[\geq 5 \] so your n must be an odd number greater than or equal to 5

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UsukiDoll
  • UsukiDoll
x, y - positive integers.. so we just have to pick all positive numbers
UsukiDoll
  • UsukiDoll
@baad1994 try letting x = 1 , y = 3, and n = 7 for x+2y=n
perl
  • perl
*
UsukiDoll
  • UsukiDoll
@perl I have no idea if I had the right idea for the first one.... T__________________T I think I did, but it's only some combinations
dumbcow
  • dumbcow
since there are an infinite num of odd integers, there will be an infinite num of (x,y) pairs i guess they want an expression in terms of n first notice that x = n - 2y , so 2y <= n-1 since x must be positive also y is pos integer 1,2,3 ... n-1 then num of possible y-values and num of pairs for any given n is (n-1)/2 total num of pairs is then just the infinite sum \[\sum_{i=2}^{\infty} \frac{n-1}{2}, n = 2i +1\] which simplifies to \[\sum_{i=2}^{\infty} i\]
perl
  • perl
\(\color{blue}{\text{Originally Posted by}}\) @dumbcow since there are an infinite num of odd integers, there will be an infinite num of (x,y) pairs i guess they want an expression in terms of n first notice that x = n - 2y , so 2y <= n-1 since x must be positive `also y is pos integer 1,2,3 ... n-1` \(\color{blue}{\text{End of Quote}}\) Question. Should that say y is a pos. integer 1,2,3,.. (n-1)/2
dumbcow
  • dumbcow
ahh yes, thank you for the correction
perl
  • perl
Nice solution :)

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