- #1
vin-math
- 10
- 0
Can anyone teach me how to find the sum of the series in terms of n in the following:
1+2x3^2+3x3^4+...+(n+1)3^2n
Thx!
1+2x3^2+3x3^4+...+(n+1)3^2n
Thx!
0rthodontist said:This can be done with generating functions if you know them.
1. Find the generating function for 1, 3^2, 3^4, ...
2. From this, find the generating function for 1, 2*3^2, ...
3. Find the generating function for (n+2)*3^(2(n+1)), (n+3)*3^(2(n+2)), ...
4. Subtract the latter from the former and evaluate at 1.
shmoe said:You might find it easier to try and sum:
[tex]1+2x^2+3x^4+\ldots+(n+1)x^{2n}[/tex]
It's not quite a geometric series, but can you turn it into one?
0rthodontist said:Okay--what sums can you do that might be relevant?
The formula for finding the sum of an arithmetic series is: S = (n/2)(a + l), where n is the number of terms, a is the first term, and l is the last term. Simply plug in these values into the formula to find the sum.
The formula for finding the sum of a geometric series is: S = a(1 - r^n) / (1 - r), where a is the first term, r is the common ratio, and n is the number of terms. Plug in these values to find the sum.
If the series has a changing pattern, it is not an arithmetic or geometric series. In this case, you can use the general formula for finding the sum of a series: S = a(1 - r^n) / (1 - r), where a is the first term, r is the common ratio, and n is the number of terms. You may need to identify the pattern and values before plugging them into the formula.
Yes, you can use a calculator to find the sum of a series. Many calculators have a "sum" function that allows you to input the series and calculate the sum directly. However, it is still important to understand the formulas and concepts behind finding the sum of a series.
A finite series has a specific number of terms and can be added up to find the sum. An infinite series has an infinite number of terms and cannot be added up to find a definite sum. However, we can use mathematical concepts such as limits to find the sum of an infinite series.