If f(x)=16x16x+4, find the value of f(1/852)+f(2/852)+f(3/852)+……..+f(851/852). A 424 B 424.5 C 425.5 D 427
solution: Check f(x)=16x16x+4 f(1−x)=161−x161−xx+4=416x+4 f(x)+f(1-x) = 16x16x+4+416x+4 = 1 f(1/852)+(851/852) = 1 f(2/852)+f(850/852) = 1 Similarly there are 425 such groups and …

Question

If f(x)=16x16x+4, find the value of f(1/852)+f(2/852)+f(3/852)+……..+f(851/852).   A 424   B 424.5   C 425.5    D 427 
solution:  Check f(x)=16x16x+4 f(1−x)=161−x161−xx+4=416x+4 f(x)+f(1-x) = 16x16x+4+416x+4 = 1 f(1/852)+(851/852) = 1 f(2/852)+f(850/852) = 1 Similarly there are 425 such groups and the middle term remains which is f(426/852) = f(1/2) = 161/2161/2+4 = 0.5 Net Sum = 425+0.5 = 425.5
can someone please explain the solution

Pv Aaditya · Answer

f(x) + f(1-x) = 1
851/852 = 1 - 1/852Similarly, 850/852 = 1 - 2/852
So, f(1/852) + f(2/852) + ... + f(851/852) = f(1/852)+f(851/852) + f(2/852)+f(850/852) + ... + f(426/852)   [Rearranging the terms]
= 1 + 1 + 1 + .... 425 times + f(426/852) = 425 + f(1/2) = 425 + 0.5 = 425.5

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