Consider a function f , satisfying f ( 1 ) + 2 f ( 2 ) + 3 f ( 3 ) + . . . + x f ( x ) = x ( x + 1 ) f ( x ) ; x 2 with f ( 1 ) = 1 . Then 1 f ( 2022 ) + 1 f ( 2028 ) is equal to [2023]

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Solution

Q.
Consider a function $f : ℕ \to ℝ,$ satisfying $f (1) + 2 f (2) + 3 f (3) + . . . + x f (x) = x (x + 1) f (x); x \geq 2$ with $f (1) = 1 .$ Then $\frac{1}{f (2022)} + \frac{1}{f (2028)}$ is equal to [2023]

1 8400

2 8200

3 8100

4 8000

Ans.
(3)

Let us consider a function $f : N \to R$ satisfying

$f (1) + 2 f (2) + 3 f (3) \dots + x f (x) = x (x + 1) f (x)$

where $x \geq 2$ with $f (1) = 1$ . We have for $x \geq 2$

$f (1) + 2 f (2) + 3 f (3) + \dots + x f (x) = x (x + 1) f (x)$

Now we will replace $x$ by $(x + 1)$ , we get

$x (x + 1) f (x) + (x + 1) f (x + 1) = (x + 1) (x + 2) f (x + 1)$

$\Rightarrow \frac{x}{f (x + 1)} + \frac{1}{f (x)} = \frac{x + 2}{f (x)} \Rightarrow x f (x) = (x + 1) f (x + 1) x \geq 2$

$f (2) = \frac{1}{4}, f (3) = \frac{1}{6}$

Now, $f (2022) = \frac{1}{4044}$

Similarly, $f (2028) = \frac{1}{4056}$

So, $\frac{1}{f (2022)} + \frac{1}{f (2028)} = 4044 + 4056 = 8100$

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