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If ƒ is a function such that ƒ(0)=2,ƒ(1)=3 and ƒ(x+2)=2ƒ(x)-ƒ(x+1) for every real x then ƒ(5) is...........
hi, in, ƒ(x+2)=2ƒ(x)-ƒ(x+1) ...................(1) put x= 0. we get ƒ(2)=2ƒ(0)-ƒ(1) = 2 *2 - 3 = 1 put x =1 , ƒ(3)=2ƒ(1)-ƒ(2) = 2* 3 - 1 = 5 put , x = 2, ƒ(4)=2ƒ(2)-ƒ(3) = 2* 1 - 5 = -3 put, x= 3 ƒ(5)=2ƒ(3)-ƒ(4) = 2*5 - (-3) = 13 ans...ƒ(5) = 13
hi,
in, ƒ(x+2)=2ƒ(x)-ƒ(x+1) ...................(1)
put x= 0. we get
ƒ(2)=2ƒ(0)-ƒ(1)
= 2 *2 - 3 = 1
put x =1 ,
ƒ(3)=2ƒ(1)-ƒ(2)
= 2* 3 - 1 = 5
put , x = 2,
ƒ(4)=2ƒ(2)-ƒ(3)
= 2* 1 - 5 = -3
put, x= 3
ƒ(5)=2ƒ(3)-ƒ(4)
= 2*5 - (-3) = 13
ans...ƒ(5) = 13
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