What is the sum when the upper bound of summation is infinite and n=1: (1-2)^n -10
*This is sigma by the way*
so it's:
∞
Sum((1-2)^n - 10
n=1
Is your(1-2)^2 supposed to be (1/2)^n ? If not, it is (-1)^n, and the series will not converge.
Why do you say n=1 in your first sentence and then talk about a summation of n from 1 to infinity later?
I am assuming that the "-10" term appears only once, AFTER the series sun, and not with each term of the series. In that case, what you have written equals -9.
The limit of the sum of (1/2)^n as n goes from 1 to infinity is 2-1 = 1
o yes sorry, it is (1/2)^n, so:
∞
Sum((1/2)^n - 10
n=1
I guess that i am saying that n starts at 1 and n goes up to infinite, so how do u put the sum of that?
As you have been told, the sum of one-half to the nth power from n equals 1 to infinity is 1. Now subtract ten.
To find the sum when the upper bound of summation is infinite, we need to evaluate the expression for each value of n and add them up. However, in this case, we notice that the term (1-2)^n will approach zero as n becomes larger and larger, since 1-2 is negative.
Therefore, the sum will converge to a constant value, rather than diverging to infinity. Let's evaluate the expression for a few values of n to see this pattern:
For n = 1, the expression is (1-2)^1 - 10 = -1 - 10 = -11.
For n = 2, the expression is (1-2)^2 - 10 = 1 - 10 = -9.
For n = 3, the expression is (1-2)^3 - 10 = -1 - 10 = -11.
As you can see, regardless of the value of n, the expression evaluates to -11 or -9, which means the sum will converge to either -11 or -9.
In conclusion, the sum when the upper bound of summation is infinite and n=1: (1-2)^n - 10 is either -11 or -9, depending on how the terms are grouped.