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Question Number 145516 by mathmax by abdo last updated on 05/Jul/21

f(x+y)=f(x)+f(y)+xy for all x and y fromR and f(4)=10 calculate f(1319)

$$\mathrm{f}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}\right)+\mathrm{xy}\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{fromR} \\ $$$$\mathrm{and}\:\mathrm{f}\left(\mathrm{4}\right)=\mathrm{10}\:\:\mathrm{calculate}\:\mathrm{f}\left(\mathrm{1319}\right) \\ $$

Answered by Olaf_Thorendsen last updated on 05/Jul/21

f(x+y) = f(x)+f(y)+xy if x = y = 2 f(4) = 2f(2)+4 ⇒ f(2) = ((10−4)/2) = 3 if x = y = 1 f(2) = 2f(1)+1 ⇒ f(1) = ((3−1)/2) = 1 if x = y = 2^(n−1) f(2^n ) = 2f(2^(n−1) )+2^(2n−2) f(2^n ) = 2(2f(2^(n−2) )+2^(2n−4) )+2^(2n−2) f(2^n ) = 2^2 f(2^(n−2) )+2^(2n−2) +2^(2n−3) ... f(2^n ) = 2^(n−2) f(2^2 )+Σ_(k=2) ^(n−1) 2^(2n−k) f(2^n ) = 10.2^(n−2) +2^(2n) Σ_(k=2) ^(n−1) ((1/2))^k f(2^n ) = 10.2^(n−2) +2^(2n) ×(1/4).((1−((1/2))^(n−2) )/(1−(1/2))) f(2^n ) = 10.2^(n−2) +2^(2n) ((1−((1/2))^(n−2) )/2) f(2^n ) = 10.2^(n−2) +2^(2n−1) −2^(n+1) 1319 = 1024+256+32+4+2+1 1319 = 2^(10) +2^8 +2^5 +4+2+1 f(1024) = 10.2^8 +2^(19) −2^(11) = 524800 f(256) = 10.2^6 +2^(15) −2^9 = 32896 f(32) = 10.2^3 +2^9 −2^6 = 528 f(1024+256) = f(1280) = f(1024)+f(256)+1024×256 f(1280) = 819840 f(1280+32) = f(1312) = f(1280)+f(32)+1280×32 f(1312) = 861328 f(1312+4) = f(1316) = f(1312)+f(4)+1312×4 f(1316) = 866586 f(1316+2) = f(1318) = f(1316)+f(2)+1316×2 f(1318) = 869221 f(1318+1) = f(1319) = f(1318)+f(1)+1318×1 f(1319) = 870540

$${f}\left({x}+{y}\right)\:=\:{f}\left({x}\right)+{f}\left({y}\right)+{xy} \\ $$$$\mathrm{if}\:{x}\:=\:{y}\:=\:\mathrm{2} \\ $$$${f}\left(\mathrm{4}\right)\:=\:\mathrm{2}{f}\left(\mathrm{2}\right)+\mathrm{4}\:\Rightarrow\:{f}\left(\mathrm{2}\right)\:=\:\frac{\mathrm{10}−\mathrm{4}}{\mathrm{2}}\:=\:\mathrm{3} \\ $$$$ \\ $$$$\mathrm{if}\:{x}\:=\:{y}\:=\:\mathrm{1} \\ $$$${f}\left(\mathrm{2}\right)\:=\:\mathrm{2}{f}\left(\mathrm{1}\right)+\mathrm{1}\:\Rightarrow\:{f}\left(\mathrm{1}\right)\:=\:\frac{\mathrm{3}−\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{1} \\ $$$$ \\ $$$$\mathrm{if}\:{x}\:=\:{y}\:=\:\mathrm{2}^{{n}−\mathrm{1}} \\ $$$${f}\left(\mathrm{2}^{{n}} \right)\:=\:\mathrm{2}{f}\left(\mathrm{2}^{{n}−\mathrm{1}} \right)+\mathrm{2}^{\mathrm{2}{n}−\mathrm{2}} \\ $$$${f}\left(\mathrm{2}^{{n}} \right)\:=\:\mathrm{2}\left(\mathrm{2}{f}\left(\mathrm{2}^{{n}−\mathrm{2}} \right)+\mathrm{2}^{\mathrm{2}{n}−\mathrm{4}} \right)+\mathrm{2}^{\mathrm{2}{n}−\mathrm{2}} \\ $$$${f}\left(\mathrm{2}^{{n}} \right)\:=\:\mathrm{2}^{\mathrm{2}} {f}\left(\mathrm{2}^{{n}−\mathrm{2}} \right)+\mathrm{2}^{\mathrm{2}{n}−\mathrm{2}} +\mathrm{2}^{\mathrm{2}{n}−\mathrm{3}} \\ $$$$... \\ $$$${f}\left(\mathrm{2}^{{n}} \right)\:=\:\mathrm{2}^{{n}−\mathrm{2}} {f}\left(\mathrm{2}^{\mathrm{2}} \right)+\underset{{k}=\mathrm{2}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{2}^{\mathrm{2}{n}−{k}} \\ $$$${f}\left(\mathrm{2}^{{n}} \right)\:=\:\mathrm{10}.\mathrm{2}^{{n}−\mathrm{2}} +\mathrm{2}^{\mathrm{2}{n}} \underset{{k}=\mathrm{2}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{k}} \\ $$$${f}\left(\mathrm{2}^{{n}} \right)\:=\:\mathrm{10}.\mathrm{2}^{{n}−\mathrm{2}} +\mathrm{2}^{\mathrm{2}{n}} ×\frac{\mathrm{1}}{\mathrm{4}}.\frac{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}−\mathrm{2}} }{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${f}\left(\mathrm{2}^{{n}} \right)\:=\:\mathrm{10}.\mathrm{2}^{{n}−\mathrm{2}} +\mathrm{2}^{\mathrm{2}{n}} \frac{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}−\mathrm{2}} }{\mathrm{2}} \\ $$$${f}\left(\mathrm{2}^{{n}} \right)\:=\:\mathrm{10}.\mathrm{2}^{{n}−\mathrm{2}} +\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} −\mathrm{2}^{{n}+\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{1319}\:=\:\mathrm{1024}+\mathrm{256}+\mathrm{32}+\mathrm{4}+\mathrm{2}+\mathrm{1} \\ $$$$\mathrm{1319}\:=\:\mathrm{2}^{\mathrm{10}} +\mathrm{2}^{\mathrm{8}} +\mathrm{2}^{\mathrm{5}} +\mathrm{4}+\mathrm{2}+\mathrm{1} \\ $$$${f}\left(\mathrm{1024}\right)\:=\:\mathrm{10}.\mathrm{2}^{\mathrm{8}} +\mathrm{2}^{\mathrm{19}} −\mathrm{2}^{\mathrm{11}} \:=\:\mathrm{524800} \\ $$$${f}\left(\mathrm{256}\right)\:=\:\mathrm{10}.\mathrm{2}^{\mathrm{6}} +\mathrm{2}^{\mathrm{15}} −\mathrm{2}^{\mathrm{9}} \:=\:\mathrm{32896} \\ $$$${f}\left(\mathrm{32}\right)\:=\:\mathrm{10}.\mathrm{2}^{\mathrm{3}} +\mathrm{2}^{\mathrm{9}} −\mathrm{2}^{\mathrm{6}} \:=\:\mathrm{528} \\ $$$$ \\ $$$${f}\left(\mathrm{1024}+\mathrm{256}\right)\:=\:{f}\left(\mathrm{1280}\right)\:=\:{f}\left(\mathrm{1024}\right)+{f}\left(\mathrm{256}\right)+\mathrm{1024}×\mathrm{256} \\ $$$${f}\left(\mathrm{1280}\right)\:=\:\mathrm{819840} \\ $$$${f}\left(\mathrm{1280}+\mathrm{32}\right)\:=\:{f}\left(\mathrm{1312}\right)\:=\:{f}\left(\mathrm{1280}\right)+{f}\left(\mathrm{32}\right)+\mathrm{1280}×\mathrm{32} \\ $$$${f}\left(\mathrm{1312}\right)\:=\:\mathrm{861328} \\ $$$${f}\left(\mathrm{1312}+\mathrm{4}\right)\:=\:{f}\left(\mathrm{1316}\right)\:=\:{f}\left(\mathrm{1312}\right)+{f}\left(\mathrm{4}\right)+\mathrm{1312}×\mathrm{4} \\ $$$${f}\left(\mathrm{1316}\right)\:=\:\mathrm{866586} \\ $$$${f}\left(\mathrm{1316}+\mathrm{2}\right)\:=\:{f}\left(\mathrm{1318}\right)\:=\:{f}\left(\mathrm{1316}\right)+{f}\left(\mathrm{2}\right)+\mathrm{1316}×\mathrm{2} \\ $$$${f}\left(\mathrm{1318}\right)\:=\:\mathrm{869221} \\ $$$${f}\left(\mathrm{1318}+\mathrm{1}\right)\:=\:{f}\left(\mathrm{1319}\right)\:=\:{f}\left(\mathrm{1318}\right)+{f}\left(\mathrm{1}\right)+\mathrm{1318}×\mathrm{1} \\ $$$${f}\left(\mathrm{1319}\right)\:=\:\mathrm{870540} \\ $$

Commented by mathmax by abdo last updated on 05/Jul/21

thanks sir olaf

$$\mathrm{thanks}\:\mathrm{sir}\:\mathrm{olaf} \\ $$

Answered by mathmax by abdo last updated on 05/Jul/21

we have f(2+2)=f(2)+f(2)+4⇒f(4)=2f(2)+4 ⇒2f(2)=10−4=6 ⇒f(2)=3 f(2+1)=f(2)+f(1)+2 =5+f(1)=f(3) f(1+1)=2f(1)+1 ⇒2f(1)=f(2)−f(1)=2 ⇒f(1)=1 ⇒f(3)=6 we have f(1)=1 ,f(2)=3=((2(2+1))/2) ,f(3)=6=((3(3+1))/2) let suppose f(n)=((n(n+1))/2) ⇒f(n+1)=f(n)+f(1)+n =((n(n+1))/2) +1+n =(n+1)((n/2)+1) =(((n+1)(n+2))/2) ⇒ ∀n ∈N^★ f(n)=((n(n+1))/2) ⇒ f(1319)=(1/2)(1319)×(1320) =660×1319=66×13190=...

$$\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{2}+\mathrm{2}\right)=\mathrm{f}\left(\mathrm{2}\right)+\mathrm{f}\left(\mathrm{2}\right)+\mathrm{4}\Rightarrow\mathrm{f}\left(\mathrm{4}\right)=\mathrm{2f}\left(\mathrm{2}\right)+\mathrm{4}\:\Rightarrow\mathrm{2f}\left(\mathrm{2}\right)=\mathrm{10}−\mathrm{4}=\mathrm{6} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{2}\right)=\mathrm{3} \\ $$$$\mathrm{f}\left(\mathrm{2}+\mathrm{1}\right)=\mathrm{f}\left(\mathrm{2}\right)+\mathrm{f}\left(\mathrm{1}\right)+\mathrm{2}\:=\mathrm{5}+\mathrm{f}\left(\mathrm{1}\right)=\mathrm{f}\left(\mathrm{3}\right) \\ $$$$\mathrm{f}\left(\mathrm{1}+\mathrm{1}\right)=\mathrm{2f}\left(\mathrm{1}\right)+\mathrm{1}\:\Rightarrow\mathrm{2f}\left(\mathrm{1}\right)=\mathrm{f}\left(\mathrm{2}\right)−\mathrm{f}\left(\mathrm{1}\right)=\mathrm{2}\:\Rightarrow\mathrm{f}\left(\mathrm{1}\right)=\mathrm{1}\:\Rightarrow\mathrm{f}\left(\mathrm{3}\right)=\mathrm{6} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{1}\right)=\mathrm{1}\:,\mathrm{f}\left(\mathrm{2}\right)=\mathrm{3}=\frac{\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right)}{\mathrm{2}}\:,\mathrm{f}\left(\mathrm{3}\right)=\mathrm{6}=\frac{\mathrm{3}\left(\mathrm{3}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\mathrm{let}\:\mathrm{suppose}\:\mathrm{f}\left(\mathrm{n}\right)=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\:\Rightarrow\mathrm{f}\left(\mathrm{n}+\mathrm{1}\right)=\mathrm{f}\left(\mathrm{n}\right)+\mathrm{f}\left(\mathrm{1}\right)+\mathrm{n} \\ $$$$=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\:+\mathrm{1}+\mathrm{n}\:=\left(\mathrm{n}+\mathrm{1}\right)\left(\frac{\mathrm{n}}{\mathrm{2}}+\mathrm{1}\right)\:=\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)}{\mathrm{2}}\:\Rightarrow \\ $$$$\forall\mathrm{n}\:\in\mathrm{N}^{\bigstar} \:\:\:\:\mathrm{f}\left(\mathrm{n}\right)=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{1319}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1319}\right)×\left(\mathrm{1320}\right)\:=\mathrm{660}×\mathrm{1319}=\mathrm{66}×\mathrm{13190}=... \\ $$

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