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```
let f : N --> R be function such that f(1) + 2f(2) + 3f(3) + ............... + nf(n) = n(n+1)f(n), for n>= 2 and f(1) = 1 then find the valueof f(n)and f(5) is ?

```
3 years ago

Meet
137 Points
```							First put the value of n=2 and then put f(1)=1 then u will get f(2)=1/4 and similarly If you put n=3 then u will get f(3)=1/6 and similarly f(4)=1/8, f(5)=1/10, so therefore by this we can say that f(n) =1/n for n>_2
```
3 years ago
mycroft holmes
272 Points
```							We have f(1)+2f(2)+...+nf(n) = n(n+1)f(n) For k=n+1 we getf(1)+2f(2)+...+n f(n)+(n+1) f(n+1) = (n+1)(n+2)f(n+1) Subtracting the two we get (n+1)(n+2) f(n+1)-n(n+1)f(n)=(n+1) f(n+1) or (n+1)(n+2) f(n+1) – (n+1) f(n+1) = n(n+1)f(n) or (n+1)2 f(n) = (n+1) f(n) or (n+1)f(n+1) = n f(n) Hence we get that for all k, kf(k) = C where C is a constant. When k=1, we see that C=1. So that f(k) = 1/k for all k
```
3 years ago
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