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Luke (Student) October 26 2015
Sheeru (Student) April 26 2015
Iftikhar (Parent) March 31 2015
Sheeru (Student) April 4 2015
Any geometric series is defined by an initial term a, and a common ratio b.
This means that we start with a and multiply by b to get each next term.
And so the general geometric series is written:
a*r^0,a*r^1,a*r^2,a*r^3,...,a*r^n ; where n+1 = no. of terms.
Now the sum of the above, S is:
S = a*r^0+a*r^1+a*r^2+a*r^3+...+ar^n
Factorise r out in most terms in right hand side:
S = a*r^0+r*(a*r^0+a*r^1+a*r^2+...+a*r^(n-1))
Now note that the sum within the brackets includes all terms in our original sum S, save the last term a*r^n. This means we can substitute (S-a*r^n) for it. Therefore,
S = a*r^0+r*(S-a*r^n)
Now all we are left to do is make the value of the sum, S, the subject of the equation:
S = a+r*S-a*r^(n+1)
S-r*S = a-a*r^(n+1)
S*(1-r) = a*(1-r*(n+1)); factorize S on LHS, and a on RHS.
S = a*(1-r*(n+1))/(1-r)