106bhaiya it is given 1f(1004)=2008
Q. Let f be a function from the set of positive integers to the set of real numbers i.e. f:N→R, such that,
(i) f(1)=1
(ii) f(1)+2f(2)+3f(3)+...+nf(n)=n(n+1)f(n) for n≥2,
then the value of 1f(1004) is _________
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5 Answers
62n(n+1)f(n)+(n+1)f(n+1)=(n+1)(n+2)f(n+1)
n(n+1)f(n)=(n+1)(n+1)f(n+1)
thus, f(n+1) = n/(n+1) f(n)
I dont knwo what mistake i am making.. but is the answer 1004?
11I am getting
\frac{1}{f(n)}=\frac{1}{\sum_{1}^{n-1}{r^{2}f(r)}}
iske baad kuch chamak ni ra
[12][12]
1Nishant you are going the right way.
f(n+1) = nn+1f(n) but this is true only for n ≥ 2
so first find f(2)
f(1) + 2f(2) = 2 x 3 x f(2)
given f(1) = 1, so we get f(2) = 1/4
so, f(n) = n-1nf(n-1) = n-1n.n-2n-1f(n-2) = .....
f(n) = n-1n.n-2n-1......23f(2)
f(n) = 2nf(2) = 2n.14 = 12n
so, 1f(1004) = 2008
62okie.. tx bipin..
even i was wondering because i was getting f(2)=4f(1)
which did not satisfy f(n+1) = n/(n+1) f(n)