The sequence {A(n)} is defined by A1=2 and A(n+1)=A(n)+2n.
what is the value of A(100)?
a)9902
b)9900
c)10100
d)9904

• Given A1 = 2
  An+1 = An + 2n
  Next number in the sequence will be,
  A2 = A1 + 2*1 which can be written as,
  A2 – A1 = 2*1  eqn.1
  A3 – A2 = 2*2  eqn.2
  A4 – A3 = 2*3  eqn.3
  and so on till,
  A100 – A99 = 2*99  eqn.99
  By adding all the above eqns. we get,
  A100 – A1 = 2*(1+2+3+4+……… + 99)
  (using sum of n integers formula = n * (n+1)/2)
  n=99
  A100 – A1 = 2 * [99 * (99 + 1)] / 2
  A100 – A1 = 99 * 100
  A100 – A1 = 9900
  A100 = 9900 + 2 = 9902
  
• 11 years agoHelpfull: Yes(117) No(12)
• Given A1 = 2
  An+1 = An + 2n
  Next number in the sequence will be,
  A1=2 or
  A2=A1+2*1=4 or 2*1+2=4
  A3=A2+2*2=8 or 3*2+2=8
  A4=A3+2*3=14 or 4*3+2=14
  A5=A4+2*4=22 or 5*4+2=22
  .......................
  simply,
  we have to find A100
  so 100*99+2=9902
• 9 years agoHelpfull: Yes(23) No(0)
• A1=2
  A2=A1+2*1=4
  A3=A2+2*2=8
  A4=A3+2*3=14
  A5=A4+2*4=22
  A6=A5+2*5=32………..A99=A98+2*98, A100=A99+2*99
  By adding all the above equations, we get
  A1+A2+……+A99+A100 = (A1+A2+…..+A99)+2(1+2+3+4+…….99)
  Cancelling L.H.S. with R.H.S.(A1 to A99 gets cancelled), We get
  A100=2*(1+2+3+…..+99)
  Using sum of n integers formula = n * (n+1)/2) where , n=99
  A100= 2 * [99 * (99 + 1)] / 2
  A100 = 99 * 100
  A100 = 9900
  Answer: 9900
• 11 years agoHelpfull: Yes(17) No(24)
• Given A1 = 2
  An+1 = An + 2n
  Next number in the sequence will be,
  A2 = A1 + 2*1 which can be written as,
  A2 – A1 = 2*1 ïƒ eqn.1
  A3 – A2 = 2*2 ïƒ eqn.2
  A4 – A3 = 2*3 ïƒ eqn.3
  and so on till,
  A100 – A99 = 2*99 ïƒ eqn.99
  By adding all the above eqns. we get,
  A100 – A1 = 2*(1+2+3+4+……… + 99)
  (using sum of n integers formula = n * (n+1)/2)
  n=99
  A100 – A1 = 2 * [99 * (99 + 1)] / 2
  A100 – A1 = 99 * 100
  A100 – A1 = 9900
  A100 = 9900 + 2 = 9902
• 11 years agoHelpfull: Yes(6) No(3)
• its b 9900
  A1 = 2
  An+1 = An + 2n
  Next number in the sequence will be,
  A2 = A1 + 2*1 which can be written as,
  A2 – A1 = 2*1  eqn.1
  A3 – A2 = 2*2  eqn.2
  A4 – A3 = 2*3  eqn.3
  and so on till,
  A100 – A99 = 2*99  eqn.99
  By adding all the above eqns. we get,
  A100= 2*(1+2+3+4+……… + 99)
  (using sum of n integers formula = n * (n+1)/2)
  n=99
  A100 = 2 * [99 * (99 + 1)] / 2
  A100 = 9900
  
• 11 years agoHelpfull: Yes(4) No(17)
• A1=2
  A2=A1+2*1=4
  A3=A2+2*2=8
  A4=A3+2*3=14
  A5=A4+2*4=22
  A6=A5+2*5=32………..A99=A98+2*98, A100=A99+2*99
  By adding all the above equations, we get
  A1+A2+……+A99+A100 = 2+((A1+A2+…..+A99)+2(1+2+3+4+…….99))
  Cancelling L.H.S. with R.H.S.(A1 to A99 gets cancelled), We get
  A100=2+2*(1+2+3+…..+99)
  Using sum of n integers formula = n * (n+1)/2) where , n=99
  A100= 2+2 * [99 * (99 + 1)] / 2
  A100 = 2+(99 * 100)
  A100 = 9902
  Answer: 9902
• 8 years agoHelpfull: Yes(1) No(0)
• A(100)=A(99)+2*99
  similarly
  A(99)=A(98)+2*98
  A(98)=A(97)+2*(97)
  :
  :
  :
  A(2)=A(1)+2*1
  so
  A(100)={2*(99+98+97+......+1)+A(1)}
  A(100)={2*(99+98+97+......+1)+2}
  A(100)=2*99/2(99+1)+2=9902
• 7 years agoHelpfull: Yes(1) No(0)
• Given A1 = 2
  An+1 = An + 2n
  
  Next number in the sequence will be,
  A2 = A1 + 2*1 which can be written as,
  A2 – A1 = 2*1 ------------- eqn.1
  A3 – A2 = 2*2 -------------- eqn.2
  A4 – A3 = 2*3 ---------------- eqn.3
  and so on till,
  
  
  A100 – A99 = 2*99 ------------------ eqn.99
  By adding all the above eqns. we get,
  A100 – A1 = 2*(1+2+3+4+……… + 99)
  (using sum of n integers formula = n * (n+1)/2)
  n=99
  A100 – A1 = 2 * [99 * (99 + 1)] / 2
  A100 – A1 = 99 * 100
  A100 – A1 = 9900
  A100 = 9900 + 2 = 9902
• 8 years agoHelpfull: Yes(0) No(0)
• A(100) = A(1) +2(1) + 2(2) + 2(3)+.................2(99)
  = 2 + 2(1+2+3+4+99)
  = 9902
  
• 7 years agoHelpfull: Yes(0) No(0)