Find the indicated sum n sigma k=1 4(0.5)^k
= 4(.5)^1 + 4(.5)^2 + 4(.5)^3 + .... + 4(.5)^n
= 4( .5^1 + .5^2 + .5^3 + ...+ .5^n)
just for the brackets,
a = .5 , r = .5 , n = n
sum(n) = .5( 1 - .5^n)/(1-.5)
= 1 - .5^n
To find the indicated sum n ∑ k=1 4(0.5)^k, we can use the formula for the sum of a geometric series.
The sum of a geometric series with a common ratio less than 1 is given by the formula:
S = a * (1 - r^n) / (1 - r)
Where:
S is the sum of the geometric series,
a is the first term of the series,
r is the common ratio, and
n is the number of terms in the series.
In this case, the first term (a) is 4, the common ratio (r) is 0.5, and we need to sum n terms, with k ranging from 1 to n.
Let's compute the sum:
n ∑ k=1 4(0.5)^k = 4 * (1 - 0.5^n) / (1 - 0.5)
So the sum of the series n ∑ k=1 4(0.5)^k is equal to 4 * (1 - 0.5^n) / (1 - 0.5).