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1.4 Sequences and Series

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Question 48

A geometric progression has first term u1 u_1\,u1​ and common ratio kkk.

Show that the sum of the first n n\,n terms of this progression, SnS_nSn​, can be expressed as

Sn=u1(1−kn)1−k S_n = \frac{u_1(1 - k^n)}{1 - k} Sn​=1−ku1​(1−kn)​
[4]

1.4 Sequences and Series Questions

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