Evaluate using the sum of a finite geometric series:
The sum of 2^i on the range from i = 1 to n. Please Help! I don't know how to make the exponent in 2^i = to i-1!
2^1 +2^1 *2^1 + 2^1* 2^2 ..... 2^1*2^(n-1)
is
2 + 2*2 + 2 *2^2 ....
note 2^1 * 2^(n-1) = 2^n
2^1 =2 = a
2 = r
sum from i = 1 to n
= 2 ( 1-2^n)/(1-2) = -2 (1-2^n) =2^(n+1) - 2
To evaluate the sum of a finite geometric series, we can use the formula:
S_n = a * (1 - r^n) / (1 - r),
where S_n is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.
In this case, the term is 2^i, and we are summing from i = 1 to n.
But the formula requires an initial term, so we need to rewrite the series so that the first term starts from i = 0. We can do this by subtracting the first term when i = 0, which is 2^0 = 1.
So, the series becomes: 2^0 + 2^1 + 2^2 + ... + 2^n.
Now we have a series that starts from i = 0. To substitute i - 1, we can rewrite the terms as follows:
2^(i - 1 + 1) = 2^i.
Now we can calculate the sum using the formula for a geometric series.
Let's put it all together:
S_n = a * (1 - r^n) / (1 - r),
where a = 1 (the first term), r = 2 (the common ratio), and n is the number of terms.
S_n = 1 * (1 - 2^n) / (1 - 2).
Simplifying further:
S_n = (1 - 2^n) / (-1).
Finally:
S_n = 2^n - 1.
Therefore, the sum of the series 2^i, for i = 1 to n, is given by 2^n - 1.