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 ?
A. 0
B. 247
C. 248
D. 249

• Solution:
  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)
• 7 years agoHelpfull: Yes(4) No(0)
• a2n=a2n-1=2^n..... now calculating 2^48
• 7 years agoHelpfull: Yes(1) No(3)
• a100= a99=2a98=a97=2^2*96 & so on... upto2^49*a0=2^49*1=2^49
  a97=2a96= ... =2^48*a0=2^48
  a96=a95=2a94= ... =2^47*a0=2^47
  SO,2^49 - 2^48 -2^47
  =2^47(4-2-1)=2^47 (ans)
• 7 years agoHelpfull: Yes(0) No(2)