Question

`(1+3+5+...+(2n-1))/(2+4+6+...+2n) = 115/116`. The value of n that satisfies is?

Answer 1

`n = 115`

Explanation:

Solution 1
Use the formula for an arithmetic series (the sum of an arithmetic sequence).

`S_n = 1/2(n(a_1+a_n))` to get

`LHS = (1/2(n(1+(2n-1))))/(1/2(n(2+2n))) = (2n)/(2+2n) = n/(1+n)`

Now solve `n/(1+n) = 115/116`

Solution 2
The sum of the first `n` odd numbers is `n^2`.

The sum of the first `n` even numbers is twice the sum of the first `n` numbers or `2((n(n+1))/2) = n(n+1)`.

So we have

`LHS = n^2/(n(n+1)) = n/(n+1)` which again leads to solving

`n/(n+1) = 115/116`

Answer 2

`n=115`

Explanation:

`(sum_(k=1)^n(2k-1))/(2 sum_(k=1)^n k) = 115/116` but

`sum_(k=1)^n(2k-1) = 2 (sum_(k=1)^n k) -n` and

`sum_(k=1)^n k= (n(n+1))/2` then

`(2(n(n+1))/2-n)/(2(n(n+1))/2) = 115/116` or

`n^2/(n(n+1))=115/116` or

`n/(n+1)=115/116 rArr n = 115`