Mark M. answered 04/17/21
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
3x / (5+x) = 3x / [5(1 - (-x/5))] = (3/5)[ x / (1 - (-x/5)]
x / (1 - (-x/5)) is the sum of the Geometric Series with first term a = x and common ratio r = -x/5.
So, we have: (3/5)∑(n=0 to infinity) arn = (3/5)∑(n=0 to infinity) x(-x/5)n = 3∑(n = 0 to infinity) [(-1)nxn+1/5n+1]
Mark M.
04/17/21
Avas A.
so to find c1, c2, c3, we should put n=1, n=2, n=3. what about "x" in the function? e.g. c1 would be: (-1)(x^2)/(5^2)04/17/21