hi
i startted the problem below by writing a sigma notation
of
sigma[(2^(n+1))/((n+1)!)]
QUESTION:
how do i find the sum of 2+(4/2!)+(8/3!)+(16/4!)+...
was my attempt wrong, or how do i go from here? thank you!
Your attempt to express the sum using sigma notation was a good start. However, we can simplify further and convert it into a more concrete expression.
We can notice that each numerator in the series is a power of 2, while each denominator is a factorial. We can rewrite the series as:
2 + (2^2)/(2!) + (2^3)/(3!) + (2^4)/(4!) + ...
To find the sum of this series, we'll break it down into separate components, calculating each term individually and then summing them up.
Term 1: 2
Term 2: (2^2)/(2!) = (4/2) = 2
Term 3: (2^3)/(3!) = (8/6) = 4/3
Term 4: (2^4)/(4!) = (16/24) = 2/3
We can see that the series has a pattern where each term is half of the previous term. So, we can generalize the nth term as (2^n)/(n!), and we can find any term by applying this formula.
To find the sum, we can sum up the individual terms as follows:
Sum = Term 1 + Term 2 + Term 3 + Term 4 + ...
Now, we can express this sum more formally using sigma notation:
Sum = Σ[(2^n)/(n!)], where n starts from 1 and goes to infinity.
Since it's not possible to calculate the infinite sum, we usually calculate an approximation by evaluating a finite number of terms.
So, to find the sum of the given series numerically, you can calculate as many terms as needed using the derived formula and then add them up.