Prove that, for a geometric series with first term (a) and common (r)
a. the sum of n term is a(r^n-1)/(r-1)
b. for I r I is less than 1, the sum of infinity is a/(1-r)
geometric series
2006-10-26 1:51 pm
回答 (2)
2006-10-26 3:45 pm
✔ 最佳答案
a) Let k = a + ar + ar^2 + ... + ar^ (n-1) ie n termsk*r = ar + ar^2 + ar^3 + ... + ar^n
If we compare the two formula above,
k*r - ar^n + a = k --- (A)
k (r-1) = a(r^n -1)
k = a(r^n -1) / (r-1)
b) Using (A) again but n is infinity here
k*r - ar^n + a = k
Because ar^n is almost zero (n is infinity)
then
k*r + a = k
k = a / (1-r)
參考: me
2006-10-26 6:42 pm
a) Let x = a + ar + ar^2 + ... + ar^ (n-1) (a geometric series with n terms) ---(1)
r*x = ar + ar^2 + ar^3 + ... + ar^n ---(2)
(2)-(1)
x(r-1) = a(r^n -1)
x = a(r^n -1) / (r-1)
b) note that x = a(1-r^n) / (1-r) for l r l < 1 --- (3)
in (3), take the limit of x (where n tends to infinity)
we know that r^n tends to 0 as n tends to infinity (for l r l < 1)
thus sum of infinity = a / (1-r)
r*x = ar + ar^2 + ar^3 + ... + ar^n ---(2)
(2)-(1)
x(r-1) = a(r^n -1)
x = a(r^n -1) / (r-1)
b) note that x = a(1-r^n) / (1-r) for l r l < 1 --- (3)
in (3), take the limit of x (where n tends to infinity)
we know that r^n tends to 0 as n tends to infinity (for l r l < 1)
thus sum of infinity = a / (1-r)
參考: myself and also maths textbook
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