If the nth partial sum of
an is sn=(3n-2)/(n+1)
what is an when x>1?
and what is the sum of an?
Well, if x > 1, then we can conclude that n+1 > 1 as well. And if we substitute that into the equation for sn, we get sn = (3n - 2) / (n + 1).
Now, to find the value of an, we just need to substitute n+1 with an in the equation. So, an = (3(an - 1) - 2) / an.
And as for the sum of an, since an is defined by a recursive formula, we can't just find a finite sum. It keeps going forever, like your never-ending to-do list. But I can guarantee you one thing, the sum will definitely give you something mind-boggling, just like trying to count all the stars in the universe!
To find the value of an when x > 1, we can substitute x = n + 1 into the given expression.
Let's plug in x = n + 1:
an = (3n - 2)/(n + 1)
= (3(n + 1) - 2)/(n + 1)
= (3n + 3 - 2)/(n + 1)
= (3n + 1)/(n + 1)
Therefore, an = (3n + 1)/(n + 1) when x > 1.
To find the sum of an, we can take the limit of the nth partial sum as n approaches infinity.
Let's evaluate the limit of sn as n approaches infinity:
Lim(sn) = Lim((3n - 2)/(n + 1)) as n approaches infinity
To find this limit, we can apply the rules of limits:
Lim(sn) = Lim((3n)/(n + 1) - (2/(n + 1)))
= Lim(3n/(n + 1)) - Lim(2/(n + 1))
As n approaches infinity, the value of 3n/(n + 1) becomes 3, and the value of 2/(n + 1) becomes 0, since the denominator is getting larger and larger. Therefore, the limit is:
Lim(sn) = 3 - 0
= 3
Hence, the sum of an is 3.
To find the value of an when x > 1, we need to substitute the given expression of sn into the formula for sn.
Given: sn = (3n - 2) / (n + 1)
First, let's express an as the difference between two consecutive partial sums:
an = sn - sn-1
Substituting the given expression for sn:
an = [(3n - 2) / (n + 1)] - [(3(n - 1) - 2) / ((n - 1) + 1)]
Simplifying the expression:
an = [(3n - 2) / (n + 1)] - (3n - 5) / n
= [(3n - 2) - (3(n - 1) - 5)] / (n + 1)
= [(3n - 2) - (3n - 3 + 5)] / (n + 1)
= (3n - 2 - 3n + 3 - 5) / (n + 1)
= (-4) / (n + 1)
= -4 / (n + 1)
Now, to find the sum of an, we substitute the formula for an into the given expression for sn:
sn = (3n - 2) / (n + 1)
= a1 + a2 + a3 + ... + an-1 + an
Substituting the expression for an:
(3n - 2) / (n + 1) = a1 + a2 + a3 + ... + an-1 + (-4) / (n + 1)
Rearranging the terms:
(3n - 2) = (a1 + a2 + a3 + ... + an-1) * (n + 1) - 4
We can simplify the right-hand side of the equation by factoring out an (n + 1) term:
(3n - 2) = [(a1 + a2 + a3 + ... + an-1) - 4 / (n + 1)] * (n + 1)
Canceling out the (n + 1) terms on both sides:
3n - 2 = a1 + a2 + a3 + ... + an-1 - 4
3n = a1 + a2 + a3 + ... + an-1 + 2
Therefore, the sum of an is:
a1 + a2 + a3 + ... + an-1 = 3n - 2 - 2
= 3n - 4
Hence, when x > 1, the value of an is -4 / (n + 1) and the sum of an is 3n - 4.
I assume that Sn = sum(1..n) An, so Sn -> 3
Dunno where the x comes from, but if you mean when n > 1,
An = Sn - s(n-1)
= (3n-2)/(n+1) - (3(n-1)-2)/(n-1+1)
= 5 / n(n+1)