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Question Number 179230 by mathlove last updated on 26/Oct/22
f(x+y)=f(x)+f(y)+x∙y and f(4)=10 faind f(2022)=?
$${f}\left({x}+{y}\right)={f}\left({x}\right)+{f}\left({y}\right)+{x}\centerdot{y} \\ $$$${and}\:{f}\left(\mathrm{4}\right)=\mathrm{10}\:\:{faind}\:\:{f}\left(\mathrm{2022}\right)=? \\ $$
Answered by mr W last updated on 26/Oct/22
let x=1, y=n−1 f(n)=f(1)+f(n−1)+n−1 f(n−1)=f(1)+f(n−2)+n−2 ..... f(3)=f(1)+f(2)+2 f(2)=f(1)+f(1)+1 Σ: f(n)=nf(1)+((n(n−1))/2) f(4)=4f(1)+((4×3)/2)=10 ⇒f(1)=1 ⇒f(n)=n+((n(n−1))/2)=((n(n+1))/2) f(2022)=((2022×(2022+1))/2)=2045253
$${let}\:{x}=\mathrm{1},\:{y}={n}−\mathrm{1} \\ $$$${f}\left({n}\right)={f}\left(\mathrm{1}\right)+{f}\left({n}−\mathrm{1}\right)+{n}−\mathrm{1} \\ $$$${f}\left({n}−\mathrm{1}\right)={f}\left(\mathrm{1}\right)+{f}\left({n}−\mathrm{2}\right)+{n}−\mathrm{2} \\ $$$$….. \\ $$$${f}\left(\mathrm{3}\right)={f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+\mathrm{2} \\ $$$${f}\left(\mathrm{2}\right)={f}\left(\mathrm{1}\right)+{f}\left(\mathrm{1}\right)+\mathrm{1} \\ $$$$\Sigma: \\ $$$${f}\left({n}\right)={nf}\left(\mathrm{1}\right)+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${f}\left(\mathrm{4}\right)=\mathrm{4}{f}\left(\mathrm{1}\right)+\frac{\mathrm{4}×\mathrm{3}}{\mathrm{2}}=\mathrm{10} \\ $$$$\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\Rightarrow{f}\left({n}\right)={n}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$${f}\left(\mathrm{2022}\right)=\frac{\mathrm{2022}×\left(\mathrm{2022}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{2045253} \\ $$
Commented by Acem last updated on 27/Oct/22
1st: Why did you do the sum? as long as f(4)=10 doesn′t give us that it is like another function which compts the descending sum of it variable. We can′t know that unless we calculate some values then we can notice that f(x+y)= g(x,y)= (((x+y)(x+y+1))/2) 2nd: 1+2+3+...+ (n−1)= ((n(n−1))/2) so what′s about sum of f(1)+f(2)+... f(n−1) 3rd: The sum you made dosen′t represents the main function. The f(4)+f(5)+...f(n) is something else... We have f(x+y)....like f(16), f(36)...etc
$$\mathrm{1}{st}:\:{Why}\:{did}\:{you}\:{do}\:{the}\:{sum}?\:{as}\:{long}\:{as}\:{f}\left(\mathrm{4}\right)=\mathrm{10} \\ $$$$\:{doesn}'{t}\:{give}\:{us}\:{that}\:{it}\:{is}\:{like}\:{another}\:{function}\: \\ $$$$\:{which}\:{compts}\:{the}\:{descending}\:{sum}\:{of}\:{it}\:{variable}. \\ $$$$\:{We}\:{can}'{t}\:{know}\:{that}\:{unless}\:{we}\:{calculate}\:{some} \\ $$$$\:{values}\:{then}\:{we}\:{can}\:{notice}\:{that} \\ $$$$\:{f}\left({x}+{y}\right)=\:{g}\left({x},{y}\right)=\:\frac{\left({x}+{y}\right)\left({x}+{y}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\: \\ $$$$\mathrm{2}{nd}:\:\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\:\left({n}−\mathrm{1}\right)=\:\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}\:{so}\:{what}'{s} \\ $$$$\:{about}\:{sum}\:{of}\:{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+…\:{f}\left({n}−\mathrm{1}\right) \\ $$$$\: \\ $$$$\mathrm{3}{rd}:\:{The}\:{sum}\:{you}\:{made}\:{dosen}'{t}\:{represents}\:{the} \\ $$$$\:\:{main}\:{function}.\:{The}\:{f}\left(\mathrm{4}\right)+{f}\left(\mathrm{5}\right)+…{f}\left({n}\right)\: \\ $$$$\:{is}\:{something}\:{else}… \\ $$$$\:{We}\:{have}\:{f}\left({x}+{y}\right)….{like}\:{f}\left(\mathrm{16}\right),\:{f}\left(\mathrm{36}\right)…{etc} \\ $$$$ \\ $$
Commented by DvMc last updated on 27/Oct/22
For your second point... I believe that in the response only was replacing the value of f(n−i) in f(n−i+1)...
$${For}\:{your}\:{second}\:{point}…\:{I}\:{believe}\:{that}\: \\ $$$${in}\:{the}\:{response}\:{only}\:{was}\:{replacing}\: \\ $$$${the}\:{value}\:{of}\:{f}\left({n}−{i}\right)\:{in}\:{f}\left({n}−{i}+\mathrm{1}\right)…\: \\ $$
Commented by mathlove last updated on 27/Oct/22
thanks
$${thanks} \\ $$
Commented by mr W last updated on 27/Oct/22
i transfered the question to following recursive sequence: a_n =a_1 +a_(n−1) +n−1 so we have a_(n−1) =a_1 +a_(n−2) +n−2 a_(n−2) =a_1 +a_(n−3) +n−3 .... a_3 =a_1 +a_2 +2 a_2 =a_1 +a_1 +1 sum all: a_n +a_(n−1) +a_(n−2) +...+a_3 +a_2 =na_1 +a_(n−1) +a_(n−2) +...+a_3 +a_2 +((n(n−1))/2) ⇒a_n =na_1 +((n(n−1))/2)
$${i}\:{transfered}\:{the}\:{question}\:{to}\:{following} \\ $$$${recursive}\:{sequence}: \\ $$$${a}_{{n}} ={a}_{\mathrm{1}} +{a}_{{n}−\mathrm{1}} +{n}−\mathrm{1} \\ $$$${so}\:{we}\:{have} \\ $$$${a}_{{n}−\mathrm{1}} ={a}_{\mathrm{1}} +{a}_{{n}−\mathrm{2}} +{n}−\mathrm{2} \\ $$$${a}_{{n}−\mathrm{2}} ={a}_{\mathrm{1}} +{a}_{{n}−\mathrm{3}} +{n}−\mathrm{3} \\ $$$$…. \\ $$$${a}_{\mathrm{3}} ={a}_{\mathrm{1}} +{a}_{\mathrm{2}} +\mathrm{2} \\ $$$${a}_{\mathrm{2}} ={a}_{\mathrm{1}} +{a}_{\mathrm{1}} +\mathrm{1} \\ $$$${sum}\:{all}: \\ $$$${a}_{{n}} +\cancel{{a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} +…+{a}_{\mathrm{3}} +{a}_{\mathrm{2}} }={na}_{\mathrm{1}} +\cancel{{a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} +…+{a}_{\mathrm{3}} +{a}_{\mathrm{2}} }+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{a}_{{n}} ={na}_{\mathrm{1}} +\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 28/Oct/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by FelipeLz last updated on 26/Oct/22
f(x+y) = f(x)+f(y)+xy x = y = a → f(2a) = 2f(a)+a^2 f(4) = 10 2f(2)+4 = 10 f(2) = 3 f(10) = f(8)+f(2)+16 = 2f(4)+f(2)+32 = 55 f(20) = 2f(10)+100 = 210 f(50) = f(40)+f(10)+400 = 2f(20)+f(10)+800 = 1275 f(100) = 2f(50)+2500 = 5050 f(200) = 2f(100)+10000 = 20100 f(500) = f(400)+f(100)+40000 = 2f(200)+f(100)+80000 = 125250 f(1000) = 2f(500)+250000 = 500500 f(2000) = 2f(1000)+1000000 = 2001000 f(2022) = f(2000)+f(22)+44000 f(2022) = 2001000+f(20)+f(2)+40+44000 f(2022) = 2001000+210+3+40+44000 f(2022) = 2045253
$${f}\left({x}+{y}\right)\:=\:{f}\left({x}\right)+{f}\left({y}\right)+{xy} \\ $$$${x}\:=\:{y}\:=\:{a}\:\rightarrow\:{f}\left(\mathrm{2}{a}\right)\:=\:\mathrm{2}{f}\left({a}\right)+{a}^{\mathrm{2}} \\ $$$$\: \\ $$$${f}\left(\mathrm{4}\right)\:=\:\mathrm{10}\: \\ $$$$\mathrm{2}{f}\left(\mathrm{2}\right)+\mathrm{4}\:=\:\mathrm{10} \\ $$$${f}\left(\mathrm{2}\right)\:=\:\mathrm{3} \\ $$$$\: \\ $$$${f}\left(\mathrm{10}\right)\:=\:{f}\left(\mathrm{8}\right)+{f}\left(\mathrm{2}\right)+\mathrm{16}\:=\:\mathrm{2}{f}\left(\mathrm{4}\right)+{f}\left(\mathrm{2}\right)+\mathrm{32}\:=\:\mathrm{55} \\ $$$${f}\left(\mathrm{20}\right)\:=\:\mathrm{2}{f}\left(\mathrm{10}\right)+\mathrm{100}\:=\:\mathrm{210} \\ $$$${f}\left(\mathrm{50}\right)\:=\:{f}\left(\mathrm{40}\right)+{f}\left(\mathrm{10}\right)+\mathrm{400}\:=\:\mathrm{2}{f}\left(\mathrm{20}\right)+{f}\left(\mathrm{10}\right)+\mathrm{800}\:=\:\mathrm{1275} \\ $$$${f}\left(\mathrm{100}\right)\:=\:\mathrm{2}{f}\left(\mathrm{50}\right)+\mathrm{2500}\:=\:\mathrm{5050} \\ $$$${f}\left(\mathrm{200}\right)\:=\:\mathrm{2}{f}\left(\mathrm{100}\right)+\mathrm{10000}\:=\:\mathrm{20100} \\ $$$${f}\left(\mathrm{500}\right)\:=\:{f}\left(\mathrm{400}\right)+{f}\left(\mathrm{100}\right)+\mathrm{40000}\:=\:\mathrm{2}{f}\left(\mathrm{200}\right)+{f}\left(\mathrm{100}\right)+\mathrm{80000}\:=\:\mathrm{125250} \\ $$$${f}\left(\mathrm{1000}\right)\:=\:\mathrm{2}{f}\left(\mathrm{500}\right)+\mathrm{250000}\:=\:\mathrm{500500} \\ $$$${f}\left(\mathrm{2000}\right)\:=\:\mathrm{2}{f}\left(\mathrm{1000}\right)+\mathrm{1000000}\:=\:\mathrm{2001000} \\ $$$$\: \\ $$$${f}\left(\mathrm{2022}\right)\:=\:{f}\left(\mathrm{2000}\right)+{f}\left(\mathrm{22}\right)+\mathrm{44000} \\ $$$${f}\left(\mathrm{2022}\right)\:=\:\mathrm{2001000}+{f}\left(\mathrm{20}\right)+{f}\left(\mathrm{2}\right)+\mathrm{40}+\mathrm{44000} \\ $$$${f}\left(\mathrm{2022}\right)\:=\:\mathrm{2001000}+\mathrm{210}+\mathrm{3}+\mathrm{40}+\mathrm{44000} \\ $$$${f}\left(\mathrm{2022}\right)\:=\:\mathrm{2045253} \\ $$
Commented by mathlove last updated on 27/Oct/22
thanks for all
$${thanks}\:{for}\:{all} \\ $$
Answered by Acem last updated on 27/Oct/22
f(4)= f(2)+f(2)+4= 10 ⇒ f(2)= 3 f(2)= f(1)+f(1)+1= 3 ⇒ f(1)= 1 f(3)= f(1)+f(2)+2= 1+3+2= 6 donc f(3)=6 f(4)= 10 We notice that the function f(x+y) represents the descending sum of the sum of it two variables i.e. x+y= 4→ 4+3 +2 +1= 10 x+y= 3→ 3+2+1=6 ....etc i.e. f(x+y)= g(x,y)= (1/2) (x+y)(x+y+1) And f(2022)= 2 045 253
$${f}\left(\mathrm{4}\right)=\:{f}\left(\mathrm{2}\right)+{f}\left(\mathrm{2}\right)+\mathrm{4}=\:\mathrm{10}\:\Rightarrow\:\boldsymbol{{f}}\left(\mathrm{2}\right)=\:\mathrm{3} \\ $$$${f}\left(\mathrm{2}\right)=\:{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{1}\right)+\mathrm{1}=\:\mathrm{3}\:\:\:\:\Rightarrow\:\boldsymbol{{f}}\left(\mathrm{1}\right)=\:\mathrm{1} \\ $$$${f}\left(\mathrm{3}\right)=\:{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+\mathrm{2}=\:\mathrm{1}+\mathrm{3}+\mathrm{2}=\:\mathrm{6}\:{donc}\:\boldsymbol{{f}}\left(\mathrm{3}\right)=\mathrm{6} \\ $$$$\:\boldsymbol{{f}}\left(\mathrm{4}\right)=\:\mathrm{10} \\ $$$$\:{We}\:{notice}\:{that}\:{the}\:{function}\:{f}\left({x}+{y}\right)\:{represents} \\ $$$$\:{the}\:{descending}\:{sum}\:{of}\:{the}\:{sum}\:{of}\:{it}\:{two}\:{variables}\:{i}.{e}. \\ $$$$\:{x}+{y}=\:\mathrm{4}\rightarrow\:\mathrm{4}+\mathrm{3}\:+\mathrm{2}\:+\mathrm{1}=\:\mathrm{10} \\ $$$$\:{x}+{y}=\:\mathrm{3}\rightarrow\:\mathrm{3}+\mathrm{2}+\mathrm{1}=\mathrm{6}\:….{etc} \\ $$$$\:{i}.{e}.\:\:{f}\left({x}+{y}\right)=\:{g}\left({x},{y}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left({x}+{y}\right)\left({x}+{y}+\mathrm{1}\right) \\ $$$$\:{And}\:\boldsymbol{{f}}\left(\mathrm{2022}\right)=\:\mathrm{2}\:\mathrm{045}\:\mathrm{253} \\ $$$$\: \\ $$

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