f-1-f-2-f-3-f-n-n-2-f-n-n-gt-1-f-1-2016-So-f-2016-

Question Number 9517 by Joel575 last updated on 12/Dec/16

f (1) + f (2) + f (3) + \dots + f (n) = n^{2} . f (n)

n > 1; f (1) = 2016

So, f (2016) = ?

Commented by sou1618 last updated on 12/Dec/16

(i) : f (1) + f (2) + \dots + f (n - 1) + f (n) = n^{2} f (n)

(i i) : f (1) + f (2) + \dots + f (n - 1) = {(n - 1)}^{2} f (n - 1)

(i) - (i i) \Rightarrow

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

\Leftrightarrow (n^{2} - 1) f (n) = {(n - 1)}^{2} f (n - 1), (n \neq 1)

\Leftrightarrow (n + 1) f (n) = (n - 1) f (n - 1), (n \neq 0)

\Leftrightarrow n (n + 1) f (n) = (n - 1) n f (n - 1)

g (n) = n (n + 1) f (n)

\Rightarrow g (n) = g (n - 1)

\Rightarrow g (n) = g (n - 1) = g (n - 2) \dots . = g (1) = 1 \times 2 \times 2016

\Rightarrow f (n) = \frac{2 \times 2016}{n (n + 1)}

f (2016) = \frac{2 \times 2016}{2016 \times 2017} = \frac{2}{2017}

Commented by Joel575 last updated on 12/Dec/16

in (i i) why u wrote {(n - 1)}^{2} f (n + 1) ? is it a typo ?

Commented by sou1618 last updated on 12/Dec/16

o h \dots . T h a n k y o u v e r y m u c h!

I f i x e d i t .

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