Consider a function f : N rightarrow R,satisf | vedclass

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(D) Given for $x \geq 2$,the sum $S_x = \sum_{k=1}^{x} k f(k) = x(x+1) f(x)$.For $x+1$,we have $S_{x+1} = S_x + (x+1) f(x+1) = (x+1)(x+2) f(x+1)$.Subs

Vedclass · Accepted Answer

$8100$

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(D) Given for $x \geq 2$,the sum $S_x = \sum_{k=1}^{x} k f(k) = x(x+1) f(x)$.For $x+1$,we have $S_{x+1} = S_x + (x+1) f(x+1) = (x+1)(x+2) f(x+1)$.Substituting $S_x = x(x+1) f(x)$ into the equation:$x(x+1) f(x) + (x+1) f(x+1) = (x+1)(x+2) f(x+1)$.Dividing by $(x+1)$ (since $x \geq 2$):$x f(x) + f(x+1) = (x+2) f(x+1)$.$x f(x) = (x+1) f(x+1)$.This implies that $n f(n)$ is a constant for $n \geq 2$.For $x=2$,$f(1) + 2 f(2) = 2(3) f(2) \Rightarrow 1 + 2 f(2) = 6 f(2) \Rightarrow 4 f(2) = 1 \Rightarrow f(2) = \frac{1}{4}$.Thus,$n f(n) = 2 f(2) = 2 	imes \frac{1}{4} = \frac{1}{2}$ for all $n \geq 2$.So,$f(n) = \frac{1}{2n}$ for $n \geq 2$.Therefore,$f(2022) = \frac{1}{2 	imes 2022} = \frac{1}{4044}$ and $f(2028) = \frac{1}{2 	imes 2028} = \frac{1}{4056}$.Finally,$\frac{1}{f(2022)} + \frac{1}{f(2028)} = 4044 + 4056 = 8100$.

Consider a function $f : N \rightarrow R$,satisfying $f(1)+2 f(2)+3 f(3)+\ldots+x f(x)=x(x+1) f(x)$ for $x \geq 2$,with $f(1)=1$. Then $\frac{1}{f(2022)}+\frac{1}{f(2028)}$ is equal to

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