# f(x)=1-2x, f:R->R Calculate the sum f(f(1))+f(f(2))+.....+f(f(10)).

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f(f(1))=1-2*f(1)=1-2*(1-2*1)

f(f(k))=1-2*f(k)=1-2*(1-2*k)=1-2+4k=4k-1

k=1 to k=10,

Sum (4k-1)=Sum (4k)- Sum 1=4Sum k-Sum 1=4*(10*11)/2 - 10=210

f(x)=1-2x. f:R->R.

Let S=Sum of ten terms of f(f(x)) , x is integers =1 to 10.

f(1)=1-2*x=-(1).f(f(1))=f(-1) =1-2*(-1)=3

f(2)=1-2*2=-3. f(f(2))=f(-3) =1-2(-3) = 7.

f(3)=1-2(3)=-5. f(f(3))=f(-5)=1-2(-5)=11.

f(x)=1-2x . f(f(x)) = f(1-2x)=1-2(1-2x) =-1+4x.

So, {-1+4x} is a sequence whose first term = 3, common difference is = 4 and the number of terms is 10.

Therefore, the sum of the first 10 terms is :

S= (First term +first term+common difference*(number of terms-1))* (number of terms)/2 = (3+3+4*(10-1))*10/2=42*10/2=**210.**