Question 569522
Suppose our sequence is a, a+d, a+2d, ..., a+2nd (this ensures that there are 2n+1 terms). Denote S_o and S_e to be the sum of the odd place terms and even place terms respectively. Then,


*[tex \LARGE S_o = a+(a+2d) + (a+4d) + ... + (a+2nd)]


*[tex \LARGE S_e = (a+d) + (a+3d) + ... + (a+(2n-1)d)]


S_o includes n+1 terms and S_e includes n terms. By equating the coefficients of a and d in each series, we have


*[tex \LARGE S_o = a(n+1) + n(n+1)d]


*[tex \LARGE S_e = an + n^2d]


Therefore, *[tex \LARGE \frac{S_o}{S_e} = \frac{a(n+1) + n(n+1)d}{an + n^2d} = \frac{(n+1)(a+nd)}{n(a+nd)} = \frac{n+1}{n}]. Multiplying both numerator and denominator by n^3 - n^2 + n - 1 yields the desired result.