Question

A function f is defined on the positive integers satisfied f(1)=2010 and f(1)+f(2) +... +f(n) =n^2f(n). Find f(4).​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:f(1) = 2010$

and

$\rm :\longmapsto\:f(1) + f(2) + - - - - + f(n) = {n}^{2}f(n)$

Now, Its a sum of n terms of series.

So, Sum of 2 terms is evaluated by substituting n = 2 in above series.

So,

$\rm :\longmapsto\:f(1) + f(2) = {2}^{2}f(2)$

$\rm :\longmapsto\:f(1) + f(2) = 4f(2)$

$\rm :\longmapsto\:f(1) = 4f(2) - f(2)$

$\rm :\longmapsto\:f(1) = 3f(2)$

$\bf\implies \:f(2) = \dfrac{f(1)}{3}$

Now, Sum of first three terms of the series is evaluated by substituting n = 3, in above given series is

$\rm :\longmapsto\:f(1) + f(2) + f(3) = {3}^{2}f(3)$

$\rm :\longmapsto\:f(1) + f(2) + f(3) = 9f(3)$

$\rm :\longmapsto\:f(1) + f(2) = 9f(3) - f(3)$

$\rm :\longmapsto\:f(1) + \dfrac{f(1)}{3} = 8f(3)$

$\rm :\longmapsto\:\dfrac{3f(1) + f(1)}{3} = 8f(3)$

$\rm :\longmapsto\:\dfrac{4f(1)}{3} = 8f(3)$

$\bf\implies \:f(3) = \dfrac{f(1)}{6}$

Now, Sum of first four terms of the series is evaluated by substituting n = 4, in above given series is

$\rm :\longmapsto\:f(1) + f(2) + f(3) + f(4) = {4}^{2}f(4)$

$\rm :\longmapsto\:f(1) + f(2) + f(3)= 16f(4) - f(4)$

$\rm :\longmapsto\:f(1) + \dfrac{f(1)}{3} + \dfrac{f(1)}{6} = 15f(4)$

$\rm :\longmapsto\:\dfrac{6f(1) + 2f(1) + f(1)}{6} = 15f(4)$

$\rm :\longmapsto\:\dfrac{9f(1)}{6} = 15f(4)$

$\rm :\longmapsto\:\dfrac{3f(1)}{2} = 15f(4)$

$\bf\implies \:f(4) = \dfrac{f(1)}{10}$

$\bf\implies \:f(4) = \dfrac{2010}{10} = 201$

Hence,

$\purple{\bf\implies \:\boxed{\tt{ \: \: \: f(4) = \dfrac{2010}{10} = 201 \: \: \: }}}$