"Consider a function $f:\mathbb{N}\to\mathbb{R}$, satisfying $f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x\ge2$ with $f(1)=1$. Then $\frac{1}{f(2022)}+\frac{1}{f(2028)}$ is equal to"

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Accepted Answer

"

$f(1) + 2f(2) + 3f(3)\, + \,...\, + \,nf(n) = n(n + 1) + (n)$ ..... (i)

$n \to n + 1$

$f(1) + 2f(2)\, + \,...\, + \,(n + 1)f(n + 1) = (n + 1)(n + 2)f(n + 1)$ ...... (ii)

(i) and (ii) gives

$3f(3) - 2f(2) = 0$

$4f(4) - 3f(3) = 0$

$\vdots$

$(n + 1)f(n + 1) - nf(n) = 0$

$\Rightarrow f(n + 1) = {{2f(2)} \over {n + 1}}$

$f(n) = {1 \over {2n}}$

${1 \over {f(2022)}} + {1 \over {f(2028)}} = 8100$

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Consider a function $f:\mathbb{N}\to\mathbb{R}$, satisfying $f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x\ge2$ with $f(1)=1$. Then $\frac{1}{f(2022)}+\frac{1}{f(2028)}$ is equal to

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Consider a function $f:\mathbb{N}\to\mathbb{R}$, satisfying $f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x\ge2$ with $f(1)=1$. Then $\frac{1}{f(2022)}+\frac{1}{f(2028)}$ is equal to

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