Given that f (x) satisfies f (x + y) = f (x) f (y) and f (1) = 1 / 2 (1), the expression of F (n) is obtained. (2) suppose an = NF (n), a1 + A2 + a3 +. + an

Given that f (x) satisfies f (x + y) = f (x) f (y) and f (1) = 1 / 2 (1), the expression of F (n) is obtained. (2) suppose an = NF (n), a1 + A2 + a3 +. + an

Let x = x, y = 1:
for:f (x+y)=f(x)f(y)
so:f(x+1)=f(x)f(1)=f(x)/2
so:f(n)=f(1)/[2^(n-1)]=1/(2^n)
a1+a2+a3+.+an=1/2+1/4+… +1/(2^n)=(2^(n+1)-1)/(2^n)=2-1/(2^n)

In the sequence {an}, an = (n-1) / N! Find the first n terms and Sn of the sequence {an}

A (n-1) = 1 / (n-2)! - 1 / (n-1)! - 1 / N! Thus, a (n-1) = 1 / (n-2)! - 1 / (n-1)! A (n-2) = 1 / (n-3)! - 1 / (n-2) A3 = 1 / 2! - 1 / 3! A2 = 1 / 1! - 1 / 2! A1 = 1 / 0! - 1 / 1! Add the above formula, and eliminate Sn = a1 + A2 + a3 + +A(n-1)+An=1/0!-1/n!=1-1/n!...

Sum: 1 + 2x + 3x ^ 2 + 4x ^ 3 + +nx^（n-1）（x≠0）
Please write down the specific steps, I want the most specific steps, I am very grateful! Hard! Urgent!

Sn=1+2x+3x^2+4x^3+…… +Nx ^ (n-1) on the other hand, write Sn = NX ^ (n-1) +... + 4x ^ 3 + 3x ^ 2 + 2x + 1, and add the two expressions to get 2Sn = n (1 + n x n-1) Sn = n (1 + n x n-1) / 2