I need to know the equations of these were n = 1
1) 1/2 - 1/4 + 1/8 - 1/16 + . . .
2) 1 + 2 + (2^2)/2 + (2^3)/6 + (2^4)/24 + (2^5)/120 + . . .
3) 0 + 3 + 8 + 15 + 24 + . . .
1/2 - 1/4 + 1/8 - 1/16 + =
1/2[1 - 1/2 + (1/2)^2 - (1/2)^3+... ]
1/2 1/(1+1/2) = 1/3
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2) exp(2)
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3) This is clearly divergent.
but what I need is actually formulas that involve n..i cant describe it
are you looking for the general term?
for 1)
1/2 - 1/4 + 1/8 - 1/16
GT = (-1)^(n+1)/2^n , where n = 1,2,3 ...
then 1/2 - 1/4 + 1/8 - 1/16 +
= [sigma] (-1)^(n+1)/2^n where n goes from 1 to ∞
let me know if this is what you are after.
Sure! Let's break down each series and find their equations.
1) 1/2 - 1/4 + 1/8 - 1/16 + ...
To find the pattern, we can observe that the denominators of the terms are powers of 2, starting from 2^0 (which is 1) and increasing by one for each subsequent term. Also, the signs alternate between positive and negative.
Using this information, we can write the general formula for the nth term (with n = 1, 2, 3, ...) as:
((-1)^(n-1)) * 1/2^n
So, when n = 1, the first term would be:
((-1)^(1-1)) * 1/2^1 = 1/2
Therefore, the equation for this series when n = 1 is simply 1/2.
2) 1 + 2 + (2^2)/2 + (2^3)/6 + (2^4)/24 + (2^5)/120 + ...
To find the pattern for this series, we can see that the terms are getting larger, but are divided by increasing factorials. In this case, the exponent of 2 is increasing by one for each term, and the denominator is the factorial of that exponent.
Using this information, the general formula for the nth term can be written as:
(2^n) / n!
So, when n = 1, the first term would be:
(2^1) / 1! = 2 / 1 = 2
Therefore, the equation for this series when n = 1 is 2.
3) 0 + 3 + 8 + 15 + 24 + ...
To find the pattern in this series, we can observe that each term is obtained by multiplying the previous term by the position of the term and then subtracting the position.
Using this information, we can write the general formula for the nth term:
n * (n + 1)
So, when n = 1, the first term would be:
1 * (1 + 1) = 1 * 2 = 2
Therefore, the equation for this series when n = 1 is 2.