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Functional feast (Posted on 2020-10-17) Difficulty: 3 of 5
Let f(x) is an odd function on R, f(1)=1 and f(x/(x-1))=xf(x) for all x<0. Find the value of

f(1)f(1/100)+f(1/2)f(1/99)+f(1/3)f(1/98)+...+f(1/50)f(1/51).

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 4.0000 (1 votes)

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Solution re: A start - a finish | Comment 2 of 3 |
(In reply to A start by Steve Herman)

V = f(1)f(1/100)+f(1/2)f(1/99)+f(1/3)f(1/98)+...+f(1/50)f(1/51) 

Lets evaluate the expression with your formulation for f(x).  Then we have 
V = (1/0!)*(1/99!)+(1/1!)*(1/98!)+(1/2!)*(1/97!)+...(1/49!)*(1/50!)

Now double V, by multiplying 2 on the left and copying the terms on the right in reverse order
2*V = [(1/0!)*(1/99!)+(1/1!)*(1/98!)+(1/2!)*(1/97!)+...(1/49!)*(1/50!)] + [(1/49!)*(1/50!) + (1/48!)*(1/51!) +...+ (1/0!)*(1/99!)]

Next, multiply both sides by 99! and simplify the fractions a bit
2*99!*V = 99!/(0!*99!)+99!/(1!*98!)+99!/(2!*97!)+...+99!/(98!*1!)+99!/(99!*0!)

At this point you may recognize the right side as row of terms in Pascal's triangle, specifically the 99th row.  So the sum is 2^99.

Then 2*99!*V = 2^99.  Therefore V = (2^98)/99!.

  Posted by Brian Smith on 2020-10-17 17:38:22
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