Maths Challenge Help: Finding 'n' When the Sum of a Series is Known (1 Viewer)

[MineTurtle]

I have a problem that I found on a practice paper for a maths challenge and would like to know how to tackle it:

Find the integral solution for 'n',
(1+3+5+7...2n-1)/(2+4+6+8...2n)=115/116

Help would be greatly appreciated!

Many Thanks,
MineTurtle

 

[Drongoski]

I have a problem that I found on a practice paper for a maths challenge and would like to know how to tackle it:

Find the integral solution for 'n',
(1+3+5+7...2n-1)/(2+4+6+8...2n)=115/116

Help would be greatly appreciated!

Many Thanks,
MineTurtle

If you know arithmetic sequences (used to be called Arithmetic Progression: AP)
we have 2 sums of an AP - one on top and one on the bottom. In this case the easiest formula for the sum of an AP is:

sum = half x number of terms x (sum of the 1st and the last term)

.: [n/2(1 + (2n-1))] / [n/2(2 + 2n)] = 115/116

Simplifying, you get: n/(n+1) = 115/116

Solving, you get: n = 115