Match the counting and series results. Column I: (A) Number of integer solutions to x + y + z = 1 with x , y , z ≥ − 4 . (B) Greatest term in 3 2 4 ( 1 + 2 1 ) 12 . (C) If a 1 , … , a 100

Question

Match the counting and series results. Column I: (A) Number of integer solutions to $x + y + z = 1$ with $x, y, z \geq - 4$ . (B) Greatest term in $\frac{4}{3 2} (1 + \frac{1}{2})^{12}$ . (C) If $a_{1}, \dots, a_{100}$ are in harmonic progression, evaluate $\sum_{i = 1}^{99} a_{i} a_{i + 1} / (a_{1} a_{100})$ . (D) Number of triangles from 11 points, eight of which are collinear. Column II: (p) 132 (q) 99 (r) 105 (s) 109

Column I	Column II
(A) Number of integer solutions to $x + y + z = 1$ with $x, y, z \geq - 4$ .	(p) 132
(B) Greatest term in $\frac{4}{3 2} (1 + \frac{1}{2})^{12}$ .	(q) 99
(C) If $a_{1}, \dots, a_{100}$ are in harmonic progression, evaluate $\sum_{i = 1}^{99} a_{i} a_{i + 1} / (a_{1} a_{100})$ .	(r) 105
(D) Number of triangles from 11 points, eight of which are collinear.	(s) 109

TestHub · Accepted Answer

A has 105 solutions after substituting u = x + 4 , v = y + 4 , w = z + 4 , so u + v + w = 13 . B's largest term is the sixth, equal to 132 . C telescopes on writing 1/ a i ​ = u + ( i − 1 ) d , so a i ​ a i + 1 ​ = ( a i ​ − a i + 1 ​ ) / d ; the ratio is ( a 1 ​ − a 100 ​ ) / ( d a 1 ​ a 100 ​ ) = 99 . D is ( 3 11 ​ ) − ( 3 8 ​ ) = 109 . Note: C maps to (q) 99; the source key misprinted C-r.

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