Let-f-N-R-be-a-function-sarisfying-following-conditions-f-1-1-f-1-2f-2-nf-n-n-n-1-f-n-Then-find-the-value-of-49f-49-

Question Number 33048 by rahul 19 last updated on 09/Apr/18

L e t f : N \to R b e a f u n c t i o n s a r i s f y i n g

f o l l o w i n g c o n d i t i o n s :

f (1) = 1 .

f (1) + 2 f (2) + \dots . + n f (n) = n (n + 1) f (n) .

T h e n f i n d t h e v a l u e o f 49 f (49) ?

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

w e l l, i c o u l d f i n d o u t t h e p a t t e r n e a s i l y

t h i s t i m e w h i c h i s \frac{1}{2 n} .

B u t h o w t o d o b y f i n d i n g g e n e r a l t e r m ?

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

i a m n o t g e t t i n g c o r r e c t a n s w e r b y

t h a t m e t h o d . a f t e r a s s u m i n g

i t t o b e {a n}^{2} + b n + c, i a m g e t t i n g

a = \frac{1}{3}, b = \frac{- 7}{4} a n d c = \frac{29}{12} .

S o m e o n e p l s c h e c k .

Commented by MJS last updated on 09/Apr/18

\dots you already found it .

the polynomial method {doesn}^{'} t

work with patterns like this,

when you already have a formular

for all n \in N

i . e .

if we know

a_{0} = 1; a_{1} = 1; a_{2} = 2; a_{3} = 6; a_{4} = 24

we believe {it}^{'} s n! but we can

find \infty functions satisfying

these pairs (n; a_{n}), one is a 4^{th} - degree

polynomial

if we know

a_{0} = 1; a_{n} = n \times a_{n - 1}

we know {it}^{'} s n!, and {there}^{'} s no

polynome for it [if you find one,

y {ou}^{'} ll get the Nobel Price; -)]

Commented by MJS last updated on 09/Apr/18

\dots just saw the other example

there we had to check the

denominator

here this leads to \frac{1}{2 n}, no more

calculation necessary

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

N o s i r, t h e t e r m s a r e

a_{1} = 1, a_{2} = \frac{1}{4}, a_{3} = \frac{1}{6}, a_{4} = \frac{1}{8} \dots \dots

\neq n!

Commented by MJS last updated on 09/Apr/18

I know, I wanted to give you

another exanple where the

polynomial approach {doesn}^{'} t

work

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

o k, t h a n k s!