Problem 16 - Entrance Test

The series is 1/2, 1/6, 1/18, 1/54, ... This is a geometric series with first term a = 1/2 and common ratio r = (1/6)/(1/2) = 1/3. Since |r| < 1, the sum to infinity is S = a/(1-r) = (1/2)/(1-1/3) = (1/2)/(2/3) = (1/2)*(3/2) = 3/4. Now compute S^2 + 2S = (3/4)^2 + 2*(3/4) = 9/16 + 6/4 = 9/16 + 24/16 = 33/16. This is not among the options. Let me recheck the series. 1/2, 1/6, 1/18, 1/54: each term is 1/3 of the previous. Yes, r=1/3. Sum = (1/2)/(1-1/3) = (1/2)/(2/3) = 3/4. S^2 + 2S = 9/16 + 3/2 = 9/16 + 24/16 = 33/16 = 2.0625. Not an integer. Perhaps the series is different. Maybe it's 1/2 + 1/4 + 1/8 + ...? No. Let me re-read: 1/2 + 1/6 + 1/18 + 1/54 + ... This is correct. S = 3/4. Then S^2 + 2S = (S)(S+2) = (3/4)(11/4) = 33/16. Not matching options. Perhaps the question asks for S^2 + 2S + 1 = (S+1)^2? (3/4+1)^2 = (7/4)^2 = 49/16. No. Maybe it's S^2 - 2S? (9/16 - 3/2) = negative. Perhaps the series sum is different. Let me compute the sum manually: 1/2 = 0.5, 1/6 ≈ 0.1667 (sum=0.6667), 1/18≈0.0556 (sum=0.7222), 1/54≈0.0185 (sum=0.7407), 1/162≈0.00617 (sum=0.7469), approaching 0.75. So S=0.75=3/4. Then S^2+2S=0.5625+1.5=2.0625. Closest option is 2 (Option B). Perhaps the intended answer is 2.