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

If f(x)=f(x−1)+x^2 +2x and f(0)=17 , find f(17).

$${If}\:\:{f}\left({x}\right)={f}\left({x}−\mathrm{1}\right)+{x}^{\mathrm{2}} +\mathrm{2}{x} \\ $$$${and}\:\:{f}\left(\mathrm{0}\right)=\mathrm{17}\:\:,\:\:{find}\:{f}\left(\mathrm{17}\right). \\ $$

Commented by ajfour last updated on 16/Jan/22

Thank you sirs.

$${Thank}\:{you}\:{sirs}. \\ $$

Answered by mahdipoor last updated on 16/Jan/22

f(1)=f(0)+1^2 +2×1 f(2)=f(1)+2^2 +2×2 f(3)=f(2)+3^2 +2×3 ... f(17)=f(16)+17^2 +2×17 ⇒f(1)+...+f(17)= f(0)+...+f(16)+(1^2 +...+17^2 )+2(1+...+17) ⇒f(17)=f(0)+((17×18×35)/6)+2((17×18)/2) ⇒f(17)=2108

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

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

f(k)−f(k−1)=(k+1)^2 −1 Σ_(k=1) ^n [f(k)−f(k−1)]=Σ_(k=1) ^n (k+1)^2 −n f(n)−f(0)=Σ_(k=2) ^(n+1) k^2 −n f(n)=f(0)+(n+1)^2 −1+((n(n+1)(2n+1))/6)−n f(n)=f(0)+((n(n+1)(2n+7))/6) f(x)=f(0)+((x(x+1)(2x+7))/6) ⇒f(x)=17+((x(x+1)(2x+7))/6) ⇒f(17)=17+((17×18×41)/6)=2108

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

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

∨ ∩i⊂∈ Sir!

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

Commented by Tawa11 last updated on 16/Jan/22

Great sirs

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

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