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Question Number 48267 by maxmathsup by imad last updated on 21/Nov/18

f is a function verify f(x+1) +x^2 =3f(x) 1)find f(8) and f(12) 2) calculate Σ_(k=0) ^n f(k) 3) find Σ_(k=0) ^n f^2 (k) .

$${f}\:{is}\:{a}\:{function}\:{verify}\:{f}\left({x}+\mathrm{1}\right)\:+{x}^{\mathrm{2}} =\mathrm{3}{f}\left({x}\right) \\ $$$$\left.\mathrm{1}\right){find}\:{f}\left(\mathrm{8}\right)\:{and}\:{f}\left(\mathrm{12}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\sum_{{k}=\mathrm{0}} ^{{n}} {f}\left({k}\right) \\ $$$$\left.\mathrm{3}\right)\:{find}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{f}^{\mathrm{2}} \left({k}\right)\:. \\ $$

Commented by maxmathsup by imad last updated on 22/Nov/18

2) we have f(x+1)−f(x)=−x^2 +2f(x) ⇒ Σ_(k=0) ^n (f(k+1)−f(k))=−Σ_(k=0) ^n k^2 +2Σ_(k=0) ^n f(k) ⇒ f(n+1)−f(0) =−((n(n+1)(2n+1))/6) +2Σ_(k=0) ^n f(k) ⇒ 2Σ_(k=0) ^n f(k)=f(n+1)−f(0)+((n(n+1)(2n+1))/6) ⇒ Σ_(k=0) ^n f(k) =(1/2){ f(n+1)−f(0)+((n(n+1)(2n+1))/6)}

$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)=−{x}^{\mathrm{2}} \:+\mathrm{2}{f}\left({x}\right)\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \left({f}\left({k}+\mathrm{1}\right)−{f}\left({k}\right)\right)=−\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}^{\mathrm{2}} \:+\mathrm{2}\sum_{{k}=\mathrm{0}} ^{{n}} {f}\left({k}\right)\:\Rightarrow \\ $$$${f}\left({n}+\mathrm{1}\right)−{f}\left(\mathrm{0}\right)\:=−\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\:+\mathrm{2}\sum_{{k}=\mathrm{0}} ^{{n}} \:{f}\left({k}\right)\:\Rightarrow \\ $$$$\mathrm{2}\sum_{{k}=\mathrm{0}} ^{{n}} {f}\left({k}\right)={f}\left({n}+\mathrm{1}\right)−{f}\left(\mathrm{0}\right)+\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\:\:\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:{f}\left({k}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{f}\left({n}+\mathrm{1}\right)−{f}\left(\mathrm{0}\right)+\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\right\} \\ $$

Commented by maxmathsup by imad last updated on 22/Nov/18

1)let suppose f polynomial ⇒f(x)=ax^2 +bx+c ⇒ a(x+1)^2 +b(x+1) +c +x^2 =3(ax^2 +bx+c) ⇒ a(x^2 +2x+1)+bx+b +c +x^2 −3ax^2 −3bx−3c =0 ⇒ ax^2 +2ax +a −2bx +(1−3a)x^2 +b−2c =0 ⇒ (1−2a)x^2 +(2a−2b)x +a+b−2c =0 ⇒ 1−2a =0 and 2a−2b =0 and a+b−2c =0 ⇒ a =(1/2) ,b=a ,c =((a+b)/2) ⇒ a=(1/2) , b=(1/2) , c=(1/2) ⇒ f(x)=(1/2) x^2 +(1/2) x+(1/2) ⇒f(8) =(8^2 /2) +(8/2) +(1/2) =32 +4 +(1/2) =36 +(1/2) =((73)/2) f(12)=((12^2 )/2) +((12)/2) +(1/2) =((144)/2) +6 +(1/2) =78 +(1/2) =((157)/2) 2) Σ_(k=0) ^n f(k) =(1/2) Σ_(k=0) ^n k^2 +(1/2) Σ_(k=0) ^n k +(1/2)Σ_(k=0) ^n (1) =(1/2) ((n(n+1)(2n+1))/6) +(1/2) ((n(n+1))/2) +(1/2)(n+1) =((n(n+1)(2n+1))/(12)) +((n(n+1))/4) +((n+1)/2) .

$$\left.\mathrm{1}\right){let}\:{suppose}\:{f}\:{polynomial}\:\Rightarrow{f}\left({x}\right)={ax}^{\mathrm{2}} \:+{bx}+{c}\:\Rightarrow \\ $$$${a}\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+{b}\left({x}+\mathrm{1}\right)\:+{c}\:+{x}^{\mathrm{2}} =\mathrm{3}\left({ax}^{\mathrm{2}} \:+{bx}+{c}\right)\:\Rightarrow \\ $$$${a}\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}\right)+{bx}+{b}\:+{c}\:+{x}^{\mathrm{2}} −\mathrm{3}{ax}^{\mathrm{2}} −\mathrm{3}{bx}−\mathrm{3}{c}\:=\mathrm{0}\:\Rightarrow \\ $$$${ax}^{\mathrm{2}} \:+\mathrm{2}{ax}\:+{a}\:\:−\mathrm{2}{bx}\:+\left(\mathrm{1}−\mathrm{3}{a}\right){x}^{\mathrm{2}} \:+{b}−\mathrm{2}{c}\:=\mathrm{0}\:\Rightarrow \\ $$$$\left(\mathrm{1}−\mathrm{2}{a}\right){x}^{\mathrm{2}} \:+\left(\mathrm{2}{a}−\mathrm{2}{b}\right){x}\:+{a}+{b}−\mathrm{2}{c}\:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{1}−\mathrm{2}{a}\:=\mathrm{0}\:{and}\:\mathrm{2}{a}−\mathrm{2}{b}\:=\mathrm{0}\:{and}\:{a}+{b}−\mathrm{2}{c}\:=\mathrm{0}\:\Rightarrow \\ $$$${a}\:=\frac{\mathrm{1}}{\mathrm{2}}\:,{b}={a}\:,{c}\:=\frac{{a}+{b}}{\mathrm{2}}\:\Rightarrow\:{a}=\frac{\mathrm{1}}{\mathrm{2}}\:,\:{b}=\frac{\mathrm{1}}{\mathrm{2}}\:,\:{c}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{f}\left(\mathrm{8}\right)\:=\frac{\mathrm{8}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{\mathrm{8}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:=\mathrm{32}\:+\mathrm{4}\:+\frac{\mathrm{1}}{\mathrm{2}}\:=\mathrm{36}\:+\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{73}}{\mathrm{2}} \\ $$$${f}\left(\mathrm{12}\right)=\frac{\mathrm{12}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{\mathrm{12}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{144}}{\mathrm{2}}\:+\mathrm{6}\:+\frac{\mathrm{1}}{\mathrm{2}}\:=\mathrm{78}\:+\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{157}}{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right)\:\sum_{{k}=\mathrm{0}} ^{{n}} {f}\left({k}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{{n}} {k}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}\:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\left({n}+\mathrm{1}\right) \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}}\:+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}}\:+\frac{{n}+\mathrm{1}}{\mathrm{2}}\:. \\ $$

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