Asked by David
So this one is confusing me:
limit of n as it approaches infinity of (n{SIGMA}i=1) (1/(2^i)
*the n is on top of the sigma notation and i=1 is at the bottom.
I understand that 'i' would equal (n(n+1))/2 however the algebra from there confuses me.
limit of n as it approaches infinity of (n{SIGMA}i=1) (1/(2^i)
*the n is on top of the sigma notation and i=1 is at the bottom.
I understand that 'i' would equal (n(n+1))/2 however the algebra from there confuses me.
Answers
Answered by
Steve
The sum is
1/(2^i)
= 1/2^1 + 1/2^2 + 1/2^3 + ...
= 1/2 + 1/4 + 1/8 + ...
You will recognize that as a geometric series, with
S = (1/2)/(1 - 1/2) = 1
1/(2^i)
= 1/2^1 + 1/2^2 + 1/2^3 + ...
= 1/2 + 1/4 + 1/8 + ...
You will recognize that as a geometric series, with
S = (1/2)/(1 - 1/2) = 1
Answered by
David
so a is the first term in the series and r is the term that is being reused over and over again ?
Answered by
David
in the equation a/(1-r)
Answered by
Steve
naturally, since that is the formula for the sum of a G.P.
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