Question

someone could help me here , please .

Answer 1

i don't know what method i have to use there. :l

Answer 2

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

Answer 3

x, y - positive integers.. so we just have to pick all positive numbers

Answer 4

@baad1994 try letting x = 1 , y = 3, and n = 7
for x+2y=n

Answer 5

*

Answer 6

@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

Answer 7

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
\[\sum_{i=2}^{\infty} i\]

Answer 8

\(\color{blue}{\text{Originally Posted by}}\) @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

Answer 9

ahh yes, thank you for the correction

Answer 10

Nice solution :)