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) ?


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

$L e t f : N \to R b e a f u n c t i o n s a r i s f y i n g$ $f o l l o w i n g c o n d i t i o n s :$ $f (1) = 1 .$ $f (1) + 2 f (2) + . . . . + n f (n) = n (n + 1) f (n) .$ $T h e n f i n d t h e v a l u e o f 49 f (49) ?$
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

$. . . you already found it .$ $the polynomial method {doesn}^{'} t$ $work with patterns like this,$ $when you already have a formular$ $for all n \in N$ $i . e .$ $if we know$ $a_{0} = 1; a_{1} = 1; a_{2} = 2; a_{3} = 6; a_{4} = 24$ $we believe {it}^{'} s n! but we can$ $find \infty functions satisfying$ $these pairs (n; a_{n}), one is a 4^{th} - degree$ $polynomial$ $if we know$ $a_{0} = 1; a_{n} = n \times a_{n - 1}$ $we know {it}^{'} s n!, and {there}^{'} s no$ $polynome for it [if you find one,$ $y {ou}^{'} ll get the Nobel Price; -)]$
Commented by MJS last updated on 09/Apr/18

$. . . just saw the other example$ $there we had to check the$ $denominator$ $here this leads to \frac{1}{2 n}, no more$ $calculation necessary$
Commented by rahul 19 last updated on 09/Apr/18

$N o s i r, t h e t e r m s a r e$ $a_{1} = 1, a_{2} = \frac{1}{4}, a_{3} = \frac{1}{6}, a_{4} = \frac{1}{8} . . . . . .$ $\neq n!$
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

$o k, t h a n k s!$
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