let a0 1 and for positive integer n define an 2an-1 if n is odd and an an-1 if n

my ans is 2^48......any1 with me???
11 years agoHelpfull: Yes(15) No(2)
a^0=1;
a^n=2a^(n-1), where n=odd;
a^n=a^(n-1), where n=even;
a^1=2a^(1-1)=2a^0=2;
a^2=a^(2-1)=a^1=2;
a^3=2a^(3-1)=2a^2=2*2=4;
a^4=a^(4-1)=a^3=4;
so,
a^100-a^97-a^96
= (a^2)^50-2a^(97-1)-a^96
=2^50- 2a^96-a^96
=2^50-3a^96
=2^2*2^48-3(a^2)^48
=4*2^48-3*2^48
=2^48(Ans)
9 years agoHelpfull: Yes(11) No(0)
cn u post a link of the image of the problem?
puzzle is not clear 2 me.
11 years agoHelpfull: Yes(4) No(0)
my ans is 2^48....any1 with me???
11 years agoHelpfull: Yes(3) No(2)
2^50-2^48-2^48
2^50-2^49
2^49{2-1}
=>2^49 ans
10 years agoHelpfull: Yes(3) No(0)
Ans is 2^48.
9 years agoHelpfull: Yes(1) No(0)
answer is 1
11 years agoHelpfull: Yes(0) No(4)
Given a0 = 1, an = 2an-1(when n is odd) and an = an-1(when n is even), then what is the value of a100-a97-a96 ?
10 years agoHelpfull: Yes(0) No(0)
2^48 is the correct answer
8 years agoHelpfull: Yes(0) No(0)
a0=1
a^1=2a0=2(1 is odd)
a^2=a^1=2(2 is even)
a^3=2a^2=4(3 is odd)
a^4=a^3=4(4 is even)
so the pattern is (a^0 to a^n): 1,2,2,4,4,8.................................................
hence a^n=2^(n/2)(when n is even)
a^n=2^[(n+1)/2](when n is odd)
hence a^100=2^50
a^97=2^49
a^96=2^48
a^100-a^97-a^96=2^50-2^49-2^48
=2^48(2^2-2-1)
=2^48(4-3)
=2^48
8 years agoHelpfull: Yes(0) No(0)

let a0=1 and for positive integer n, define an=2an-1 if n is odd; and an=an-1 if n is even, Then what is the value of a100-a97-a96 equals

a0=a not plz understand clearly i can't type correctly in the puzzle box