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1.6 Sequences and Series

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Question 51

The first three terms of a geometric series are (2k+2)(2k+2)(2k+2), (k+4)(k+4)(k+4), and k k\,k respectively, where k k\,k is a positive constant.

a.

Show that k2−6k−16=0k^2 - 6k - 16 = 0k2−6k−16=0.

[4]
b.

Hence show that k=8k = 8k=8.

[2]
c.

Find the common ratio.

[2]
d.

Find the sum to infinity of the series.

[2]

1.6 Sequences and Series Questions

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