Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 71666 by aliesam last updated on 18/Oct/19

f:z→z    f(x+y)=f(x)+f(y)+3(4xy−1)    ,f(1)=0    ∀x,y ∈z  evaluate f(19)

f:zzf(x+y)=f(x)+f(y)+3(4xy1),f(1)=0x,yzevaluatef(19)

Commented by kaivan.ahmadi last updated on 18/Oct/19

f(2)=f(1+1)=9  f(3)=f(1+2)=0+9+21=30  f(6)=f(3+3)=30+30+135=195  f(4)=f(2+2)=9+9+45=63  f(10)=f(6+4)=195+63+285=543  f(9)=f(3+6)=30+195+213=438  f(19)=f(10+9)=543+438+1077=2058

f(2)=f(1+1)=9f(3)=f(1+2)=0+9+21=30f(6)=f(3+3)=30+30+135=195f(4)=f(2+2)=9+9+45=63f(10)=f(6+4)=195+63+285=543f(9)=f(3+6)=30+195+213=438f(19)=f(10+9)=543+438+1077=2058

Answered by mind is power last updated on 18/Oct/19

f(x+1)=f(x)+3(4x−3)=f(x)+12x−3  ⇒f(x+1)−f(x)=12x−3  ∀x∈Z  ⇒Σ_(x=1) ^(18) [f(x+1)−f(x)]=Σ_(x=1) ^(18) (12x−9)  ⇒f(19)−f(1)=((12+12.18)/2).18−3.18  ⇒f(19)=9(12+216)−54=1998

f(x+1)=f(x)+3(4x3)=f(x)+12x3f(x+1)f(x)=12x3xZ18x=1[f(x+1)f(x)]=18x=1(12x9)f(19)f(1)=12+12.182.183.18f(19)=9(12+216)54=1998

Commented by aliesam last updated on 18/Oct/19

thank you sir nice sol

thankyousirnicesol

Commented by mind is power last updated on 18/Oct/19

y′re welcom

yrewelcom

Terms of Service

Privacy Policy

Contact: info@tinkutara.com