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Question Number 164396 by ajfour last updated on 16/Jan/22

$$Iff\left(x\right)=f(x-1)+x^2+2x$$$$andf\left(0\right)=17,findf\left(17\right).$$

Commented by ajfour last updated on 16/Jan/22

$$Thankyousirs.$$

Answered by mahdipoor last updated on 16/Jan/22

$$f\left(1\right)=f\left(0\right)+1^2+2\times 1$$$$f\left(2\right)=f\left(1\right)+2^2+2\times 2$$$$f\left(3\right)=f\left(2\right)+3^2+2\times 3$$$$...$$$$f\left(17\right)=f\left(16\right)+{17}^2+2\times 17$$$$\Rightarrow f\left(1\right)+...+f\left(17\right)=$$$$f\left(0\right)+...+f\left(16\right)+(1^2+...+{17}^2)+2(1+...+17)$$$$\Rightarrow f\left(17\right)=f\left(0\right)+\frac{17\times 18\times 35}{6}+2\frac{17\times 18}{2}$$$$\Rightarrow f\left(17\right)=2108$$

Answered by mr W last updated on 16/Jan/22

$$f\left(k\right)-f(k-1)={(k+1)}^2-1$$$$\underset{k=1}{\sum ^n}[f\left(k\right)-f(k-1)]=\underset{k=1}{\sum ^n}{(k+1)}^2-n$$$$f\left(n\right)-f\left(0\right)=\underset{k=2}{\sum ^{n+1}}{k}^2-n$$$$f\left(n\right)=f\left(0\right)+{(n+1)}^2-1+\frac{n(n+1)(2n+1)}{6}-n$$$$f\left(n\right)=f\left(0\right)+\frac{n(n+1)(2n+7)}{6}$$$$f\left(x\right)=f\left(0\right)+\frac{x(x+1)(2x+7)}{6}$$$$\Rightarrow f\left(x\right)=17+\frac{x(x+1)(2x+7)}{6}$$$$\Rightarrow f\left(17\right)=17+\frac{17\times 18\times 41}{6}=2108$$

Commented by Rasheed.Sindhi last updated on 16/Jan/22

$$\vee \cap \mathbf{i}\subset \in \mathrm{Sir}!$$

Commented by Tawa11 last updated on 16/Jan/22

$$\mathrm{Great}\mathrm{sirs}$$

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