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

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) ?

$${Let}\:{f}:{N}\rightarrow{R}\:{be}\:{a}\:{function}\:{sarisfying} \\ $$$${following}\:{conditions}: \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}. \\ $$$${f}\left(\mathrm{1}\right)+\mathrm{2}{f}\left(\mathrm{2}\right)+....+{nf}\left({n}\right)={n}\left({n}+\mathrm{1}\right){f}\left({n}\right). \\ $$$${Then}\:{find}\:{the}\:{value}\:{of}\:\mathrm{49}{f}\left(\mathrm{49}\right)\:? \\ $$

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

well, i could find out the pattern easily  this time which is (1/(2n)).  But how to do by finding general term?

$${well},\:{i}\:{could}\:{find}\:{out}\:{the}\:{pattern}\:{easily} \\ $$$${this}\:{time}\:{which}\:{is}\:\frac{\mathrm{1}}{\mathrm{2}{n}}. \\ $$$${But}\:{how}\:{to}\:{do}\:{by}\:{finding}\:{general}\:{term}? \\ $$

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

i am not getting correct answer by  that method. after assuming   it to be an^2 +bn+c , i am getting  a=(1/3), b= ((−7)/4) and c= ((29)/(12)).  Someone pls check.

$${i}\:{am}\:{not}\:{getting}\:{correct}\:{answer}\:{by} \\ $$$${that}\:{method}.\:{after}\:{assuming}\: \\ $$$${it}\:{to}\:{be}\:{an}^{\mathrm{2}} +{bn}+{c}\:,\:{i}\:{am}\:{getting} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{3}},\:{b}=\:\frac{−\mathrm{7}}{\mathrm{4}}\:{and}\:{c}=\:\frac{\mathrm{29}}{\mathrm{12}}. \\ $$$${Someone}\:{pls}\:{check}. \\ $$

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∈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 ∞ functions satisfying  these pairs (n;a_n ), one is a 4^(th) −degree  polynomial  if we know  a_0 =1; a_n =n×a_(n−1)   we know it′s n!, and there′s no  polynome for it [if you find one,  you′ll get the Nobel Price ;−)]

$$...\mathrm{you}\:\mathrm{already}\:\mathrm{found}\:\mathrm{it}. \\ $$$$\mathrm{the}\:\mathrm{polynomial}\:\mathrm{method}\:\mathrm{doesn}'\mathrm{t} \\ $$$$\mathrm{work}\:\mathrm{with}\:\mathrm{patterns}\:\mathrm{like}\:\mathrm{this}, \\ $$$$\mathrm{when}\:\mathrm{you}\:\mathrm{already}\:\mathrm{have}\:\mathrm{a}\:\mathrm{formular} \\ $$$$\mathrm{for}\:\mathrm{all}\:{n}\in\mathbb{N} \\ $$$$\mathrm{i}.\mathrm{e}. \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{know} \\ $$$${a}_{\mathrm{0}} =\mathrm{1};\:{a}_{\mathrm{1}} =\mathrm{1};\:{a}_{\mathrm{2}} =\mathrm{2};\:{a}_{\mathrm{3}} =\mathrm{6};\:{a}_{\mathrm{4}} =\mathrm{24} \\ $$$$\mathrm{we}\:\boldsymbol{\mathrm{believe}}\:\mathrm{it}'\mathrm{s}\:{n}!\:\boldsymbol{\mathrm{but}}\:\mathrm{we}\:\mathrm{can} \\ $$$$\mathrm{find}\:\infty\:\mathrm{functions}\:\mathrm{satisfying} \\ $$$$\mathrm{these}\:\mathrm{pairs}\:\left({n};{a}_{{n}} \right),\:\mathrm{one}\:\mathrm{is}\:\mathrm{a}\:\mathrm{4}^{\mathrm{th}} −\mathrm{degree} \\ $$$$\mathrm{polynomial} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{know} \\ $$$${a}_{\mathrm{0}} =\mathrm{1};\:{a}_{{n}} ={n}×{a}_{{n}−\mathrm{1}} \\ $$$$\mathrm{we}\:\boldsymbol{\mathrm{know}}\:\mathrm{it}'\mathrm{s}\:{n}!,\:\mathrm{and}\:\mathrm{there}'\mathrm{s}\:\mathrm{no} \\ $$$$\mathrm{polynome}\:\mathrm{for}\:\mathrm{it}\:\left[\mathrm{if}\:\mathrm{you}\:\mathrm{find}\:\mathrm{one},\right. \\ $$$$\left.\mathrm{y}\left.\mathrm{ou}'\mathrm{ll}\:\mathrm{get}\:\mathrm{the}\:\mathrm{Nobel}\:\mathrm{Price}\:;−\right)\right] \\ $$

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 (1/(2n)), no more  calculation necessary

$$...\mathrm{just}\:\mathrm{saw}\:\mathrm{the}\:\mathrm{other}\:\mathrm{example} \\ $$$$\mathrm{there}\:\mathrm{we}\:\mathrm{had}\:\mathrm{to}\:\mathrm{check}\:\mathrm{the} \\ $$$$\mathrm{denominator} \\ $$$$\mathrm{here}\:\mathrm{this}\:\mathrm{leads}\:\mathrm{to}\:\frac{\mathrm{1}}{\mathrm{2}{n}},\:\mathrm{no}\:\mathrm{more} \\ $$$$\mathrm{calculation}\:\mathrm{necessary} \\ $$

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

No sir, the terms are  a_1 = 1 , a_2 = (1/4) , a_3 = (1/6) , a_4 =(1/8) ......  ≠ n!

$${No}\:{sir},\:{the}\:{terms}\:{are} \\ $$$${a}_{\mathrm{1}} =\:\mathrm{1}\:,\:{a}_{\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{4}}\:,\:{a}_{\mathrm{3}} =\:\frac{\mathrm{1}}{\mathrm{6}}\:,\:{a}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{8}}\:...... \\ $$$$\neq\:{n}! \\ $$

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

$$\mathrm{I}\:\mathrm{know},\:\mathrm{I}\:\mathrm{wanted}\:\mathrm{to}\:\mathrm{give}\:\mathrm{you} \\ $$$$\mathrm{another}\:\mathrm{exanple}\:\mathrm{where}\:\mathrm{the} \\ $$$$\mathrm{polynomial}\:\mathrm{approach}\:\mathrm{doesn}'\mathrm{t} \\ $$$$\mathrm{work} \\ $$

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

ok, thanks!

$${ok},\:{thanks}! \\ $$

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