a-function-f-is-defined-on-the-positive-integers-satisfies-f-1-1002-amp-f-1-f-2-f-3-f-n-n-2-f-n-find-f-2003-

Question Number 90259 by john santu last updated on 22/Apr/20

a f u n c t i o n f i s d e f i n e d o n

t h e p o s i t i v e i n t e g e r s s a t i s f i e s

f (1) = 1002, & f (1) + f (2) + f (3) +

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

f i n d f (2003)

Answered by mr W last updated on 22/Apr/20

\underset{k = 1}{\sum^{n}} f (k) = n^{2} f (n) \dots (i)

\underset{k = 1}{\sum^{n + 1}} f (k) = {(n + 1)}^{2} f (n + 1)

\underset{k = 1}{\sum^{n}} f (k) + f (n + 1) = {(n + 1)}^{2} f (n + 1) \dots (i i)

(i i) - (i) :

f (n + 1) = {(n + 1)}^{2} f (n + 1) - n^{2} f (n)

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

\Rightarrow f (n) = \frac{n - 1}{n + 1} f (n - 1)

= \frac{(n - 1) (n - 2) \dots 2 \times 1}{(n + 1) n (n - 1) \dots 4 \times 3} f (1)

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

f (2003) = \frac{2}{2004 \times 2003} \times 1002 = \frac{1}{2003}

Commented by john santu last updated on 22/Apr/20

t h e r e l a t i o n y o u o b t a i n e d b e t w e e n

f (n) a n d f (n - 1) i s c o r r e c t .

b y t h e w a y, i n t h e s t e p b e f o r e

w i t h f (n + 1) y o u w r o t e f (n - 1)

i n s t e a d o f f (n) . t h e r e l a t i o n

y o u o b t a i n e d f (n) a n d f (1) i s

i n c o r r e c t . i t s h o u l d b e a 2 a n d

n o t 6 . t h e q u o t i e n t o f p r o d u c t

{d o e s n}^{'} t s e e m r i g h t . s o i^{'} m

a s s u m i n g t h e e r r o r c a m e

f r o m t h e r e .

Commented by john santu last updated on 22/Apr/20

t h e a n s w e r i s \frac{1}{2003}

Commented by mr W last updated on 22/Apr/20

n o w f i x e d . t h a n k s!

Facebook Tweet Pin