The first three terms of a geometric series are (2k−2)(2k-2)(2k−2), (k+3)(k+3)(k+3), and k k\,k respectively, where k k\,k is a positive constant.
Show that k2−8k−9=0k^2 - 8k - 9 = 0k2−8k−9=0.
Hence show that k=9k = 9k=9.
Find the common ratio.
The sum to infinity of the series.
