既知の関数f(x)f(x+y)=f(x)f(y)かつf(1)=1/2(1)f(n)を求める式(2)an=nf(n)をa1+a2+a3+.+an

既知の関数f(x)f(x+y)=f(x)f(y)かつf(1)=1/2(1)f(n)を求める式(2)an=nf(n)をa1+a2+a3+.+an

x=x,y=1:
for:f(x+y)=f(x)f(y)
so:f(x+1)=f(x)f(1)=f(x+1)/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)