The first three terms of a geometric series are (3k+3)(3k+3)(3k+3), (k+3)(k+3)(k+3), and k k\,k respectively, where k k\,k is a positive constant.
Show that 2k2−3k−9=02k^2 - 3k - 9 = 02k2−3k−9=0.
Hence show that k=3k = 3k=3.
Find the common ratio.
Find the sum to infinity of the series.
