the series 1/(2^n) from n=0 .... i need to know at what (n) the sum will be about 10... looking for an answer and explaination.. or at least tell me what method i can use...
2006-10-26 19:13:19 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The series can be written as: 1, 1/2, 1/(2^2), 1/(2^3), ...... ,1/(2^n), .... to infinity.
This is an infinite Geometric Series.
If we consider the first sum of first 'n'-terms,
Sn = [1 - (1/2)^n] / [1 - (1/2)] = 2*[1 - (1/2)^n]
As you say for Sn = 10 we have to find 'n'.
Put Sn = 10 in the above equation
10 = 2*[1 - (1/2)^n]
5 = [1 - (1/2)^n]
(1/2)^n = -4
Right hand side is a negative quantity and left hand side is a positive quantity.
This is impossible.
So for no value of 'n' the sum can be equal to 10.
2006-10-26 19:22:53 · answer #1 · answered by psbhowmick 6 · 1⤊ 0⤋
Sum[1/(2^n)] from n = 0 to infinity converges to 2.
The series above is a convergent geometric series with r = 1/2.
It converges because r<1. It converges to 1/(1-r), which in this case is just 2.
2006-10-27 02:23:09 · answer #2 · answered by Gypsy Catcher 3 · 1⤊ 0⤋
Are you sure the question is written correctly, because as is, 1 + 1/2 +1/4 + 1/8 etc. will never have sum = 10. (In fact, it will approach 2 as n approaches infinity)
2006-10-27 02:17:26 · answer #3 · answered by topher8128 2 · 1⤊ 0⤋
the sum of 1/2^n = 1.1111111 ( binary notation )
thius sum will never be 10 for some n.
2006-10-27 02:17:30 · answer #4 · answered by gjmb1960 7 · 1⤊ 0⤋