Show the sum from n=0 to 200 of x^n given that x is not 1, is (1-x^201)/(1-x) hence find the sum of 1+2(1/2)+3(1/2)^2+...+200(1/2)^199

Let the first term of a geometric progression be 'A', define the geometric sum, which we will call 'S', of 'Ax^n' for each integer n to be the sum from n=0 to r of 'Ax^n'. Hence S=A+Ax+Ax^2+...+Ax^r. Consider the product Sx=Ax+Ax^2+...+Ax^(r+1). By subtracting Sx from S we get: S-Sx=A-Ax^(r+1)=A(1-x^(r+1))=S(1-x). Given that x does not equal 1, and by dividing by (1-x) we deduce that the sum S=A(1-x^(r+1))/(1-x). The first term 'A' of x^n is x^0=1, by observing r=200 we deduce that S for the sum in the question is (1-x^201)/(1-x)By looking at the sum 1+2(1/2)+...+199(1/2)^200 we see that each term can be expressed as nx^(n-1) for x=1/2. By observing that nx^(n-1) is the derivative of x^n we conclude that we are required to differentiate the sum from n=0 to 200 of x^n.This gives: sum from n=0 to 200 of nx^(n-1)=d/dx((1-x^201)/(1-x))=1/(1-x)*((1-x^201)/(1-x)-201x^200) which can be done via chain rule. For x=1/2, this gives an answer of approximately 4.