S=1+2(1/5)+3(1/5)^2+4(1/5)^3….
If S=a/b, where a and b are co prime positive integers, what is the value of a+b?
S(x) = 1 + x + x^2 + x^3 +... = 1/(1-x)
dS/dx = 1 + 2 x + 3 x^2 + 4 x^3 +...
= 1/(1-x)^2
what's the sum of infinite terms?
thanks i have now understood the problem.............
41
To find the value of S, we can start by noticing that the given series resembles a geometric series.
A geometric series has the form:
S = a + ar + ar^2 + ar^3 + ...
Where a is the first term, r is the common ratio, and S is the sum of the series.
In the given series, the first term a is 1, and the common ratio r is (1/5).
So, the given series can be rewritten as:
S = 1 + 2(1/5) + 3(1/5)^2 + 4(1/5)^3 + ...
Now, we can multiply the entire series by (1/5) to obtain a series with a common ratio of (1/5) (same as the previous series):
(1/5)S = (1/5) + 2(1/5)^2 + 3(1/5)^3 + 4(1/5)^4 + ...
Next, we subtract this series from the original series to eliminate the coefficients:
S - (1/5)S = 1 + (1/5) + (1/5)^2 + (1/5)^3 + ...
We can see that the right side of the equation now represents an infinite geometric series with the first term a = 1 and the common ratio r = (1/5).
Using the formula to find the sum of an infinite geometric series, which is S = a / (1 - r), we can substitute the values:
S - (1/5)S = 1 / (1 - 1/5)
Simplifying the right side, we have:
(4/5)S = 1 / (4/5)
To isolate S, we divide both sides by (4/5):
S = (1 / (4/5)) * (5/4)
S = (5/4) * (5/4)
S = 25 / 16
Therefore, S = 25 / 16.
To find the value of a + b, we need to determine the numerator and the denominator.
The numerator is 25, and the denominator is 16.
Since 25 and 16 have no common factors other than 1, they are co-prime.
So, a = 25 and b = 16.
Finally, to find a + b:
a + b = 25 + 16 = 41.
Therefore, the value of a + b is 41.