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A function $f(x) $ satisfies $f(1)=3600$ and $f(1)+f(2)+ ... f(n) =n^2 f(n)$, for all positive integers n>1. What is the value of $f(9)$?

  1. 80
  2. 240
  3. 200
  4. 100
  5. 120
asked in Quantitative Aptitude by (7.1k points)   | 53 views

1 Answer

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Best answer
f1+f2=4f2

f2=f1/3

f1+f2+f3=9f3

f1+f1/3+f3=9f3

f3=f1/6

f1+f2+f3+f4=16f4

We get f4=f1/10

So we can say fn=f1/sum of n terms

therefore f9=3600/45=80
answered by (1.6k points)  
edited by
@pooja : Answer should be 80

             & that should be : fn=f1/sum.of n terms.
f9 = f1/45 = 3600/45 = 80

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