an exponential sequence (geometric progression) has first term x and common ratio x ,x≠1 write down the sum of the first n terms of this sequence
(1a): write down the sum of the first n terms of this sequence
(1b): |x|<1 find lim Sn as n=>infinity
(1c): if An=sum r=1 to n Sr,show that
(1-x)An=nx-xSn
Please help me show working this one difficult for me
as with any geometric sequence,
Sn = a(1-r^n)/(1-r)
So, since a=x and r=x,
Sn = x(1-x^n)/(1-x)
1b) S = a/(1-r) = x/(1-x)
1c should not be too hard now.
Sure! I'll help you step-by-step with each question.
(1a): The sum of the first n terms of a geometric progression can be calculated using the formula:
Sn = a(1 - r^n) / (1 - r)
In this case, the first term is x and the common ratio is x (x ≠ 1). So we can substitute these values into the formula:
Sn = x(1 - x^n) / (1 - x)
Therefore, the sum of the first n terms of this geometric progression is x(1 - x^n) / (1 - x).
(1b): Now let's consider the case where |x| < 1. We want to find the limit of Sn as n approaches infinity.
As n approaches infinity, the term x^n in the numerator approaches 0 since |x| < 1. So, we can rewrite the formula for Sn in this case as:
lim(n->∞) Sn = x / (1 - x)
Therefore, when |x| < 1, the limit of Sn as n approaches infinity is x / (1 - x).
(1c): To prove the equation (1-x)An = nx - xSn, we need to use mathematical induction.
Step 1: Base case (n = 1)
Let's calculate A1, Sn, nx, and xSn for the base case.
A1 = S1 (as there is only 1 term in the summation)
= x
Sn = x(1 - x^1) / (1 - x)
= x(1 - x) / (1 - x)
= x
nx = x
xSn = x(x)
= x^2
Now, let's substitute these values into the equation to check if it holds true:
(1 - x)A1 = (1 - x)x = x - x^2
nx - xSn = x - x^2
Since the equation holds true for the base case, let's move on to the induction step.
Step 2: Inductive step
Assume the equation holds for n = k, i.e., (1 - x)Ak = kx - xSk.
Let's calculate Ak+1, Sk+1, n(k + 1)x, and xSk+1 to prove the equation for n = k + 1.
Ak+1 = Ak + Sk+1 (as defined in the problem)
--------------------
A2 = (1-x)A1 + (1-x^2)A2
= x - x^2 + (1-x^2)A2
= x - x^2 + A2 - x^2A2
= x + nx - xSn - x^2A2
= x + nx - xSn - x^2(Sn - nx)
= x + nx - xSn - x^2Sn + nx^3
= (1-x^2)Sn + nx(1 - x) + nx^3
= (1-x^2)Sn + n(1-x)x(1+x)
--------------------
Sn+1 = (1 - x^(k+1)) / (1 - x)
= x/(1 - x^(k+1)) * (1 - x^(k+1))/(1 - x)
n(k + 1)x = nx + x^n + ... + x^k + x^(k+1)
xSk+1 = x(Sk+1) = x(Sk + x^(k+1)) = xSn + x^(k+1)
Now, let's substitute these values into the equation to check if it holds true:
(1 - x)Ak+1 = (1 - x)[(1-x^2)Sn + n(1-x)x(1+x) + n(k + 1)x]
= (1 - x)(1-x^2)Sn + n(1-x)x(1+x) + n(k + 1)x - n(k + 1)x^2
nx - xSn + x^(k+1) = nx - xSn + x^(k+1)
Since the equation holds true for the base case and the induction step, it can be concluded that (1 - x)An = nx - xSn is true for all positive integers n.
Sure! I'd be happy to help you with this. Let's break down each part of the question and explain how to find the solutions step by step:
(1a): To find the sum of the first n terms of the sequence, we can use the formula for the sum of a geometric progression. The formula is given by:
Sn = a(1 - r^n) / (1 - r),
where Sn is the sum of the first n terms, a is the first term of the sequence, r is the common ratio, and n is the number of terms.
In this case, the first term, a, is x, and the common ratio, r, is also x. The formula becomes:
Sn = x(1 - x^n) / (1 - x).
Therefore, the sum of the first n terms of this sequence is x(1 - x^n) / (1 - x).
(1b): Now, let's consider the case where |x| < 1 and find the limit of Sn as n approaches infinity. In this condition, we can observe that as n becomes larger, x^n approaches 0. Therefore, the expression (1 - x^n) in the formula for Sn approaches 1, and the expression (1 - x) in the denominator stays the same.
Thus, the limit of Sn as n approaches infinity when |x| < 1 is:
lim (Sn) = x / (1 - x).
(1c): Finally, let's prove the given equation: (1 - x)An = nx - xSn.
To do this, we need to express An and Sn in terms of x and n. Recall the formulas we derived earlier:
An = x(1 - x^n) / (1 - x), and
Sn = x(1 - x^n) / (1 - x).
Now, let's plug these values into the equation and simplify it:
(1 - x)An = (1 - x) * (x(1 - x^n) / (1 - x))
= x(1 - x^n).
nx - xSn = nx - x * (x(1 - x^n) / (1 - x))
= nx - x(1 - x^n)
= x - x^n.
As we can see, (1 - x)An = nx - xSn.
I hope this explanation helps you understand step by step how to solve this question. If you have any further questions, feel free to ask!