Suppose that Sn = 3 + 1/3 + 1/9 + ... + 1/3(n-2) a) Find S10 and S∞ b) If the common difference in an arithmetic sequence is twice the first term, show that Sn/Sm = n^2/m^2 a) Sum of the geometric sequence is a = 3 and r = 1/3 (a(1 - r)^n)/(1 - r) (3(1 - 1/3)^9)/(1 - 1/3) S10 = 4.499771376 For infinity, as n goes to infinity, the numerator goes to 1 so we have S∞ = 3(1)/2/3 = 4.5 b) Sum of an arithmetic sequence formula is below: n(a1 + an)/2 an = a1 + (n - 1)2a1 since d = 2a1 n(a1 + a1 + (n - 1)2a1)/2 (2a1n + n^2 - 2a1n)/2 n^2/2 For Sm m(a1 + am)/2 am = a1 + (m - 1)2a1 since d = 2a1 m(a1 + 1 + (m - 1)2a1)/2 (2a1m + m^2 - 2a1m)/2 m^2/2 Sn/Sm = n^2/m^2 (cancel the 2's) S10/S1 = 10^2/1^2 We know S<sub>1</sub> = 3 So we have 100(3)/1 S10 = 300