The series (4x-3)+(4x-3)^2+(4x-3)^3................................is given for which values of X will the series converge??
To determine the values of x for which the series converges, we can use the concept of convergence of a series. In this case, we can analyze the series by examining the behavior of the terms as n approaches infinity.
The given series is (4x-3) + (4x-3)^2 + (4x-3)^3 + ...
Let's analyze the general term of the series, which can be expressed as (4x-3)^n.
For the series to converge, the general term must approach zero as n approaches infinity. So, we need to find the conditions under which (4x-3)^n approaches zero as n gets larger.
Consider the term (4x-3)^n.
For (4x-3)^n to approach zero, we need to ensure that the absolute value of (4x-3) is less than 1.
|4x-3| < 1
Now, we can solve this inequality to find the values of x for which the series converges.
We have two cases:
Case 1: (4x-3) > 0
If (4x-3) > 0, then the inequality becomes:
4x-3 < 1
Adding 3 to both sides:
4x < 4
Dividing both sides by 4:
x < 1
Case 2: (4x-3) < 0
If (4x-3) < 0, then the inequality becomes:
-(4x-3) < 1
Multiplying both sides by -1 (keep in mind that it reverses the inequality direction):
4x-3 > -1
Adding 3 to both sides:
4x > 2
Dividing both sides by 4:
x > 0.5
Therefore, combining the results from both cases, we can conclude that the series will converge for x values in the following range:
0.5 < x < 1
In summary, the given series will converge for x values between 0.5 and 1.