Hi. I'm having trouble with the summation of n from i=2 of ((2^n-i)(2^n)). Any help would be greatly appreciated!
the way I am reading the question you have i is going from 2 to n
so there will be n-1 terms
using i = 2,3,4,..,n
I get 2^(2n-2) + 2^(2n-2) + ...
Sum of n terms of a geometric series is
a(1 - r^n)/(1-r)
so a= 2^(2n-2), r = 2^(-1) and the number of terms is n-1
and the sum is
2^(2n-2)[1 - 2^(-1)] / 1-2^(-1))
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=2^(n-1)(2^n - 1)
To sum the expression ((2^(n-i))(2^n)) from i = 2 to n, you can use the formula for the sum of a geometric series.
First, let's determine the common ratio of the series. In this case, the common ratio is 2^(-1) because each term is obtained by dividing the previous term by 2.
Next, let's determine the first term of the series. When i = 2, the first term is (2^(n-2))(2^n).
Now, using the formula for the sum of a geometric series, we have:
Sum = (first term - last term) / (1 - common ratio)
The last term is obtained when i = n, which gives us (2^(n-n))(2^n) = 2^n.
Plugging these values into the formula, we get:
Sum = ((2^(n-2))(2^n) - 2^n) / (1 - 2^(-1))
= (2^(n-2+1))(2^n - 1) / (1 - 1/2)
= 2^(n-1)(2^n - 1)
So the sum of the given expression is 2^(n-1)(2^n - 1).