Master the digits!
Calculus
Level
pending
If f(x) be a function such that f(x-1)+f(x+1)=√3 f(x) and f(5)=10, then the sum of digits of the value of Σf(5+12 r) from r=0 to r=19 is
The answer is 2.
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1 solution
Given: f(x-1) +f(x+1) = √3 f(x)
f(x+2) +f(x)= √3 f(x+1)
Putting x=x+2,
f(x+1)+ f(x+3)= f(x+2)
f(x-1)+2 f(x+2)+ f(x+3)= √3[√3f(x+1)]
f(x-1)+ f(x+3)= f(x+1)
Putting x=x+2 again
f(x+1)+ f(x+5)= f(x+3)
f(x-1)+ f(x+5)=0
f(x+5)=- f(x+1)
f(x)=- f(x+6)
f(x+12)= f(x)
∑f(5+12r) from r=0 tor=19 is 20f(5)=20*10=200
Hence, sum of digits is 2+0+0=2