Evaluate each telescoping sum:
1) Sum from i = 1 to 100 of (5^i - 5^(i-1)).
I know that I could apply the (n(n+1))/2 summation rule to the 5^i part of the question and then plug in 100, but I am unsure what to do for the second half --> -5^(i-1)! Thank you for any help!
How about just writing down a few terms?
5^1 - 5^0
5^2 - 5^1
...
5^100 - 5^99
---------------
5^100 - 5^0
If you want to do it the hard way, you could note that
∑(a-b) = ∑a - ∑b
To evaluate the telescoping sum from i = 1 to 100 of (5^i - 5^(i-1)), we can simplify the expression and then apply the telescoping sum technique.
Let's start by simplifying the expression:
5^i - 5^(i-1) = 5^i - 5^(i-1)
= 5^i - (1/5) * 5^i
= 5^i * (1 - 1/5)
= 5^i * (4/5)
= (4/5) * 5^i
Now, let's express the telescoping sum:
S = (4/5) * 5^1 + (4/5) * 5^2 + (4/5) * 5^3 + ... + (4/5) * 5^100
To evaluate this sum, we can factor out the common term (4/5) and write it as:
S = (4/5) * (5^1 + 5^2 + 5^3 + ... + 5^100)
Now, let's try to find a pattern in the sum by examining a few terms:
The sum of the first two terms is 5^1 + 5^2, which can be written as:
5^1 + 5^2 = 5^1 * (1 + 5) = 6 * 5^1
The sum of the first three terms is 5^1 + 5^2 + 5^3:
5^1 + 5^2 + 5^3 = 5^1 * (1 + 5 + 25) = 31 * 5^1
Looking at the pattern, we can observe that the sum of the first n terms is:
5^1 + 5^2 + 5^3 + ... + 5^n = 5^1 * (1 + 5 + 5^2 + ... + 5^(n-1))
Using the formula for the sum of a geometric series, we can simplify this expression:
5^1 * (1 + 5 + 5^2 + ... + 5^(n-1)) = 5^1 * ((5^n - 1)/(5 - 1)) = 5^(n+1) - 5
Now, let's substitute this expression back into our sum S:
S = (4/5) * (5^1 * (1 + 5 + 5^2 + ... + 5^100))
= (4/5) * (5^(100+1) - 5)
= (4/5) * (5^101 - 5)
And there you have the evaluation of the telescoping sum from i = 1 to 100 of (5^i - 5^(i-1)).
To get the final numerical value, you can simply calculate (4/5) * (5^101 - 5) using a calculator.
To evaluate the telescoping sum from i = 1 to 100 of (5^i - 5^(i-1)), we can begin by simplifying the expression:
5^i - 5^(i-1) = 5^i - (5^i * 5^(-1)) = 5^i - (5^i / 5) = 5^i - (1/5) * 5^i.
Now, let's rewrite the sum using this new expression:
Sum from i = 1 to 100 of (5^i - 5^(i-1)) = Sum from i = 1 to 100 of (5^i - (1/5) * 5^i).
Next, we can factor out the common term 5^i from each term in the sum:
= Sum from i = 1 to 100 of (1 - (1/5)) * 5^i
= (1 - 1/5) * (Sum from i = 1 to 100 of 5^i)
The sum from i = 1 to 100 of 5^i can be evaluated using the formula for the sum of a geometric series:
Sum from i = 1 to n of r^i = (r^(n+1) - r) / (r - 1),
where n is the number of terms and r is the common ratio.
So, using this formula, the sum from i = 1 to 100 of 5^i can be written as:
(5^(100+1) - 5) / (5 - 1) = (5^101 - 5) / 4.
Finally, substituting this result back into the expression:
= (1 - 1/5) * (5^101 - 5) / 4
= 4/5 * (5^101 - 5) / 4
= (5^101 - 5) / 5.
Therefore, the evaluation of the telescoping sum is (5^101 - 5) / 5.