f-N-R-f-1-2005-and-f-1-f-2-f-n-n-2-f-n-n-gt-1-Then-f-2004-

Question Number 33032 by rahul 19 last updated on 09/Apr/18

f : N \to R

f (1) = 2005 .

a n d

f (1) + f (2) + \dots \dots + f (n) = n^{2} f (n), n > 1 .

T h e n f (2004) = ?

Commented by Joel578 last updated on 09/Apr/18

The denomitator form a series

\underset{⌣}{1 3} \underset{⌣}{6} \underset{⌣}{10} \underset{⌣}{15} \underset{⌣}{21}, \dots

+ 2 + 3 + 4 + 5 + 6

Suppose the general term T_{n} = {a n}^{2} + b n + c

T_{1} = a + b + c = 1

T_{2} = 4 a + 2 b + c = 3

T_{3} = 9 a + 3 b + c = 6

After few calculations, you will found

a = \frac{1}{2}, b = \frac{1}{2}, c = 0

\to T_{n} = \frac{n^{2}}{2} + \frac{n}{2} = \frac{n}{2} (n + 1)

Commented by Joel578 last updated on 09/Apr/18

Sir, I think the equation is

f (1) + f (2) + \dots + f (n) = n^{2} f (n)

Commented by rahul 19 last updated on 09/Apr/18

J o e l s i r, s u p p o s e w e a r e n o t a b l e t o

s e e t h e p a t t e r n a t t h e f i r s t s i g h t .

T h e n h o w c a n w e f i n d t h e g e n e r a l

t e r m ?

Commented by Rasheed.Sindhi last updated on 09/Apr/18

Sorry sir I misunderstood then .

Commented by rahul 19 last updated on 09/Apr/18

t h a n k u s o m u c h s i r!

Answered by Joel578 last updated on 09/Apr/18

f (1) = \frac{2005}{1}

f (2) = \frac{2005}{3}

f (3) = \frac{2005}{6}

f (4) = \frac{2005}{10}

f (5) = \frac{2005}{15}

⋮

f (n) = \frac{2005}{\frac{n}{2} (n + 1)}

f (2004) = \frac{2005}{1002 . 2005} = \frac{1}{1002}

Commented by rahul 19 last updated on 09/Apr/18

t h a n k u s i r!