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Question Number 87352 by jagoll last updated on 04/Apr/20

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(\mathrm{x}+1)-\mathrm{x}$$$$\mathrm{f}\left(0\right)=123$$$$\mathrm{f}\left(29\right)=?$$

Commented by jagoll last updated on 04/Apr/20

$$\mathrm{x}=0\Rightarrow 123=\mathrm{f}\left(1\right)-0$$$$\mathrm{f}\left(1\right)=123$$$$\mathrm{x}=1\Rightarrow 123=\mathrm{f}\left(2\right)-1\Rightarrow \mathrm{f}\left(2\right)=124$$$$\mathrm{x}=2\Rightarrow 124=\mathrm{f}\left(3\right)-2\Rightarrow \mathrm{f}\left(3\right)=126$$$$\mathrm{x}=3\Rightarrow 126=\mathrm{f}\left(4\right)-3\Rightarrow \mathrm{f}\left(4\right)=129$$$$\mathrm{it}\mathrm{correct}$$

Commented by jagoll last updated on 04/Apr/20

$$\mathrm{f}\left(29\right)=123+\frac{29\times 28}{2}$$$$=123+406=529$$

Commented by jagoll last updated on 04/Apr/20

$$\mathrm{mr}\mathrm{W},\mathrm{my}\mathrm{work}\mathrm{it}\mathrm{correct}?$$

Commented by mr W last updated on 04/Apr/20

$$yes$$

Commented by john santu last updated on 04/Apr/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}$$$$\mathrm{f}(\mathrm{x}+1)=\mathrm{a}{(\mathrm{x}+1)}^2+\mathrm{b}(\mathrm{x}+1)+\mathrm{c}$$$$\mathrm{f}(\mathrm{x}+1)-\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}$$$$2\mathrm{ax}+\mathrm{a}+\mathrm{b}=\mathrm{x}$$$$2\mathrm{a}=1\Rightarrow \mathrm{a}=\frac{1}{2}$$$$\mathrm{a}+\mathrm{b}=0\Rightarrow \mathrm{b}=-\frac{1}{2}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}{\mathrm{x}}^2-\frac{1}{2}\mathrm{x}+\mathrm{c}$$$$\mathrm{f}\left(0\right)=\mathrm{c}=123$$$$\therefore \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}{\mathrm{x}}^2-\frac{1}{2}\mathrm{x}+123$$$$\mathrm{then}\mathrm{f}\left(29\right)=\frac{1}{2}\times 29(29-1)+123$$$$\frac{29\times 28}{2}+123$$

Commented by jagoll last updated on 04/Apr/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by mr W last updated on 04/Apr/20

$$f(k+1)=f\left(k\right)+k$$$$\underset{k=0}{\sum ^n}f(k+1)=\underset{k=0}{\sum ^n}f\left(k\right)+\underset{k=0}{\sum ^n}k$$$$f(n+1)-f\left(0\right)+\underset{k=0}{\sum ^n}f\left(k\right)=\underset{k=0}{\sum ^n}f\left(k\right)+\underset{k=0}{\sum ^n}k$$$$f(n+1)-f\left(0\right)=\underset{k=0}{\sum ^n}k=\frac{n(n+1)}{2}$$$$f(n+1)=f\left(0\right)+\frac{n(n+1)}{2}$$$$\Rightarrow f\left(n\right)=f\left(0\right)+\frac{n(n-1)}{2}$$$$\Rightarrow f\left(29\right)=123+\frac{29\times 28}{2}=529$$

Commented by jagoll last updated on 04/Apr/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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