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)231 exam-style questions on WJEC A Level Maths 3.4 Sequences and Series (A-level only), covering 3.4.1 Sequences and Series (A-level only), 3.4.2 Sequences and Series (A-level only), 3.4.3 Sequences and Series (A-level only), 3.4.4 Sequences and Series (A-level only), 3.4.5 Sequences and Series (A-level only), 3.4.6 Sequences and Series (A-level only), and 3.4.7 Sequences and Series (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.