Respuesta :
Uh! That's really hard. There is no closed formula. There are few cases besides arithmetic and geometric that can be in fact summed up. For instance srithmetico-geometric. But there is no general formula.
Let's use an example to demonstrate this point.
Assume we want to find the sum of: 1 + 3 + 7 + 15 + 31 and so on.
This is clearly neither geometric nor is it arithmetic.
We can further express this in sigma notation, and this will come in handy with what we want to do next.
[tex]T(n) = \sum_{n = 1}^{\infty}(2^{n} - 1)[/tex]
Clearly, this is true to our summation.
If we plug n = 1, we have 2 - 1 = 1.
If we plug n = 2, we have 1 + (4 - 1) = 1 + 3.
Now, using sigma laws, we can express this as BOTH, a geometric and an arithmetic sequence.
[tex]T(n) = \sum_{n = 1}^{\infty}(2^{n} - 1) = \sum_{n = 1}^{\infty}2^{n} - \sum_{n = 1}^{\infty}1[/tex]
From here, we can use our formulae to solve this.
To summarise: Look for any way to split up the pattern into both sequences.
Assume we want to find the sum of: 1 + 3 + 7 + 15 + 31 and so on.
This is clearly neither geometric nor is it arithmetic.
We can further express this in sigma notation, and this will come in handy with what we want to do next.
[tex]T(n) = \sum_{n = 1}^{\infty}(2^{n} - 1)[/tex]
Clearly, this is true to our summation.
If we plug n = 1, we have 2 - 1 = 1.
If we plug n = 2, we have 1 + (4 - 1) = 1 + 3.
Now, using sigma laws, we can express this as BOTH, a geometric and an arithmetic sequence.
[tex]T(n) = \sum_{n = 1}^{\infty}(2^{n} - 1) = \sum_{n = 1}^{\infty}2^{n} - \sum_{n = 1}^{\infty}1[/tex]
From here, we can use our formulae to solve this.
To summarise: Look for any way to split up the pattern into both sequences.
