Question

More @jim_thompson5910 sorry

Answer 1

\[\frac{ 1+3+5+...+(2n-1) }{ 2+4+6+...+2n} = \frac{ 115 }{ 116 }\]

Answer 2

find a positive integral solution for n

Answer 3

choices: 110, 115, 116, 231, or no answer

Answer 4

again I'm going to use

1+2+3+...+n = n*(n+1)/2

Answer 5

1+3+5+...+(2n-1) = 2*1-1+2*2-1+2*3-1+...+2*n-1
1+3+5+...+(2n-1) = (2*1+2*2+2*3+...+2*n) + (-1 - 1 - 1...-1)
1+3+5+...+(2n-1) = 2*(1+2+3+...+n) + (-1 - 1 - 1...-1)
1+3+5+...+(2n-1) = 2*n*(n+1)/2 + n*(-1)
1+3+5+...+(2n-1) = n*(n+1) - n


2+4+6+...+2n = 2*(1+2+3+...+n)
2+4+6+...+2n = 2*n*(n+1)/2
2+4+6+...+2n = n*(n+1)

Answer 6

the numerator simplifies to n*(n+1) - n

the denominator simplifies to n*(n+1)

Answer 7

and I guess you could factor out n to get


n*(n+1) - n = n[(n+1) - 1]
n*(n+1) - n = n[n+1 - 1]
n*(n+1) - n = n[n+0]
n*(n+1) - n = n*n
n*(n+1) - n = n^2

Answer 8

It is an arithmetic progression divided by another arithmetic progression.

Answer 9

\[\large \frac{ 1+3+5+...+(2n-1) }{ 2+4+6+...+2n} = \frac{ 115 }{ 116 }\]

turns into

\[\large \frac{ n^2 }{ n*(n+1)} = \frac{ 115 }{ 116 }\]

which simplifies to

\[\large \frac{ n }{ n+1} = \frac{ 115 }{ 116 }\]

Answer 10

So you could use: \[
\sum a_n=\frac n2 (a_i+a_n)
\]

Answer 11

yes that's probably a much faster way

Answer 12

so its 115

Answer 13

yeah