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Question Number 199547 by hardmath last updated on 05/Nov/23

Answered by mr W last updated on 05/Nov/23

1) n×n!=(n+1−1)×n!=(n+1)!−n! 1×1!+2×2!+...+2022×2022! =2!−1!+3!−2!+....+2023!−2022! =2023!−1! ⇒ab=2023×1=2023 2) f(a+b)=f(a)+f(b)+ab f(2a)=2f(a)+a^2 f(2)=2f(1)+1^2 =5 ⇒f(1)=2 f(4)=2f(2)+2^2 =2×5+4=14 f(5)=f(4)+f(1)+4×1=14+2+4=20 f(10)=2f(5)+5^2 =2×20+25=65 f(11)=f(10)+f(1)+10×1=65+2+10=77 or generally f(x)=((x(x+3))/2) ⇒f(11)=((11×14)/2)=77

$$\left.\mathrm{1}\right) \\ $$$${n}×{n}!=\left({n}+\mathrm{1}−\mathrm{1}\right)×{n}!=\left({n}+\mathrm{1}\right)!−{n}! \\ $$$$\mathrm{1}×\mathrm{1}!+\mathrm{2}×\mathrm{2}!+...+\mathrm{2022}×\mathrm{2022}! \\ $$$$=\mathrm{2}!−\mathrm{1}!+\mathrm{3}!−\mathrm{2}!+....+\mathrm{2023}!−\mathrm{2022}! \\ $$$$=\mathrm{2023}!−\mathrm{1}! \\ $$$$\Rightarrow{ab}=\mathrm{2023}×\mathrm{1}=\mathrm{2023} \\ $$$$ \\ $$$$\left.\mathrm{2}\right) \\ $$$${f}\left({a}+{b}\right)={f}\left({a}\right)+{f}\left({b}\right)+{ab} \\ $$$${f}\left(\mathrm{2}{a}\right)=\mathrm{2}{f}\left({a}\right)+{a}^{\mathrm{2}} \\ $$$$ \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{2}{f}\left(\mathrm{1}\right)+\mathrm{1}^{\mathrm{2}} =\mathrm{5}\:\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{2} \\ $$$${f}\left(\mathrm{4}\right)=\mathrm{2}{f}\left(\mathrm{2}\right)+\mathrm{2}^{\mathrm{2}} =\mathrm{2}×\mathrm{5}+\mathrm{4}=\mathrm{14} \\ $$$${f}\left(\mathrm{5}\right)={f}\left(\mathrm{4}\right)+{f}\left(\mathrm{1}\right)+\mathrm{4}×\mathrm{1}=\mathrm{14}+\mathrm{2}+\mathrm{4}=\mathrm{20} \\ $$$${f}\left(\mathrm{10}\right)=\mathrm{2}{f}\left(\mathrm{5}\right)+\mathrm{5}^{\mathrm{2}} =\mathrm{2}×\mathrm{20}+\mathrm{25}=\mathrm{65} \\ $$$${f}\left(\mathrm{11}\right)={f}\left(\mathrm{10}\right)+{f}\left(\mathrm{1}\right)+\mathrm{10}×\mathrm{1}=\mathrm{65}+\mathrm{2}+\mathrm{10}=\mathrm{77} \\ $$$$ \\ $$$${or}\:{generally}\:{f}\left({x}\right)=\frac{{x}\left({x}+\mathrm{3}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{f}\left(\mathrm{11}\right)=\frac{\mathrm{11}×\mathrm{14}}{\mathrm{2}}=\mathrm{77} \\ $$

Commented by hardmath last updated on 05/Nov/23

thank you dear professor cool

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool} \\ $$

Commented by hardmath last updated on 05/Nov/23

thank you dear professor perfect solutions

$${thank}\:{you}\:{dear}\:{professor}\:{perfect}\:{solutions} \\ $$

Commented by hardmath last updated on 05/Nov/23

dear professor, how did you get the generalized solution and is it always true?

$$\mathrm{dear}\:\mathrm{professor}, \\ $$$$\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{the}\:\mathrm{generalized}\:\mathrm{solution} \\ $$$$\mathrm{and}\:\mathrm{is}\:\mathrm{it}\:\mathrm{always}\:\mathrm{true}? \\ $$

Commented by mr W last updated on 05/Nov/23

since f(x+y)=f(x)+f(y)+xy, you can assume f(x)=ax^2 +bx+c with f(2)=5. then you get a=(1/2), b=(3/2), c=0.

$${since}\:{f}\left({x}+{y}\right)={f}\left({x}\right)+{f}\left({y}\right)+{xy},\:{you}\:{can} \\ $$$${assume}\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}\:{with}\:{f}\left(\mathrm{2}\right)=\mathrm{5}. \\ $$$${then}\:{you}\:{get}\:{a}=\frac{\mathrm{1}}{\mathrm{2}},\:{b}=\frac{\mathrm{3}}{\mathrm{2}},\:{c}=\mathrm{0}. \\ $$

Answered by deleteduser1 last updated on 05/Nov/23

f(4)=10+4=14⇒f(8)=44 5=2f(1)+1⇒f(1)=2;f(3)=7+2=9 f(11)=44+9+24=77

$${f}\left(\mathrm{4}\right)=\mathrm{10}+\mathrm{4}=\mathrm{14}\Rightarrow{f}\left(\mathrm{8}\right)=\mathrm{44} \\ $$$$\mathrm{5}=\mathrm{2}{f}\left(\mathrm{1}\right)+\mathrm{1}\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{2};{f}\left(\mathrm{3}\right)=\mathrm{7}+\mathrm{2}=\mathrm{9} \\ $$$${f}\left(\mathrm{11}\right)=\mathrm{44}+\mathrm{9}+\mathrm{24}=\mathrm{77} \\ $$

Commented by hardmath last updated on 05/Nov/23

thank you professor

$${thank}\:{you}\:{professor} \\ $$

Answered by ajfour last updated on 05/Nov/23

2. f(x+h)−f(x)=f(h)+hx f ′(x)=((f(0))/h)+x ⇒ f(0)=0 f(x)=(x^2 /2)+kx as f(2)=(2^2 /2)+2k=5 ⇒ k=(3/2) f(x)=(x^2 /2)+((3x)/2)=((x(x+3))/2)

$$\mathrm{2}.\:\:{f}\left({x}+{h}\right)−{f}\left({x}\right)={f}\left({h}\right)+{hx} \\ $$$${f}\:'\left({x}\right)=\frac{{f}\left(\mathrm{0}\right)}{{h}}+{x} \\ $$$$\Rightarrow\:\:{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{kx} \\ $$$${as}\:\:\:{f}\left(\mathrm{2}\right)=\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}{k}=\mathrm{5}\:\:\:\:\Rightarrow\:\:{k}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{3}{x}}{\mathrm{2}}=\frac{{x}\left({x}+\mathrm{3}\right)}{\mathrm{2}} \\ $$

Commented by hardmath last updated on 05/Nov/23

thank you professor

$${thank}\:{you}\:{professor} \\ $$

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