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Difficult G.P. Problem from a Difficult Math Book

  1. Apr 3, 2014 #1
    1. The problem statement, all variables and given/known data
    15. The series $$a+ar-ar^2-ar^3+ar^4+ar^5-ar^6-ar^7+...,$$
    where ##a>0##, has its ##k##th term, ##T_k##, defined by
    $$T_k=ar^{k-1}$$ if ##k## is of the form ##4p-3## or ##4p-2## and
    $$T_k=-ar^{k-1}$$ if ##k## is of the form ##4p-1## or ##4p,##
    where p is a positive integer. By rewriting the series as the sum of two geometric series, or otherwise, prove that the sum to ##4n## terms of the series is $$\frac{a(1+r)(1-r^4n)}{(1+r^2)}.$$

    2. Relevant equations
    $$S_n=\frac{a(1-r^n)}{1-r}$$


    3. The attempt at a solution
    $$a+ar-ar^2-ar^3+ar^4+ar^5-ar^6-ar^7+...$$
    $$=(a+ar+ar^4+ar^5+...)-(ar^2+ar^3+ar^6+ar^7+...)$$
    $$=(a+ar^4+ar^8+...)+(ar+ar^5+ar^9+...)-(ar^2+ar^6+ar^10+...)-(ar^3+ar^7+ar^{11}+...)$$
    $$={}^1 S_n +{}^2 S_n +{}^3 S_n +{}^4 S_n$$
    $$=\frac{a(1-(r^{4n})}{1-r^4}+\frac{ar(1-(r^{4n})}{1-r^4}-\frac{ar^2(1-(r^{4n})}{1-r^4}-\frac{ar^3(1-(r^{4n})}{1-r^4}$$
    $$=a(1+r+r^2+r^3)\frac{a(1-r^{4n})}{(1-r^4)}$$
    and now I am stUCK in the mUCK:yuck: YUCK! I don't know how to factorise ##1+r+r^2+r^3##.
     
    Last edited: Apr 3, 2014
  2. jcsd
  3. Apr 3, 2014 #2
    Try grouping terms like this: (1 + r) + (r2 + r3). 1 + r is then a common factor of both terms.
     
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