53) The sequence {An } is defined by A1 = 2 and An+1 =An + 2n. What is the value of A100 ?
a) 9902
b) 9900
c) 10100
d) 9904

• ans is 9902
  a1=2,An+1=An+2n
  from this we can get series 2,4,8,14,22,32,....,
  here 4-2=2,8-4=4,14-8=6,22-14=8,so difference is an a.p.
  and 1st term + sum of differences=last term.
  so use 2 as first term and 99 terms are there
  sum of series of difference=(99/2)(2*2+(99-1)2)(A.P. formula to find sum of n terms)
  hence A100=A1+sum of series of difference=2+9900=9902.
• 11 years agoHelpfull: Yes(41) No(4)
• Ans : a)9902
  From given,An = An-1 + 2(n-1)
  since A1=2,An = 2 + 2*1 + 2*2 +2*3 +......2*(n-1)
  Taking 2 common,An = 2*(1 + (1+2+3+......n-1) )
  Now,1+2+3+.....n is given by n*(n+1)/2.
  Therefore,1+2+3....n-1 = (n-1)*n/2
  Therefore, An = 2*(1+n(n-1)/2) = 2+n(n-1) => An = 2+n(n-1)
  Hence, A100 = 2+100*99 = 9902
• 11 years agoHelpfull: Yes(6) No(2)
• ans c)10100
  see the sequence:
  A(1)=2, A(2)=6, A(3)=12, A(4)=20... and so on..
  the pattern we can see is
  A(2)=2*(2+1)
  A(3)=3*(3+1) and so on.. so
  A(100)=100*(100+1)=101*100=10100
  
  so ans is c
• 11 years agoHelpfull: Yes(2) No(6)
• please apply the concept of recursion:
  A(n+1)=A(n)+2n
  ={A(n-1)+2(n-1)}+2n
  ={A(n-2}+2(n-2)}+2(n-1)+2n
  =A(n-2)+2{(n-2)+(n-1)+n)}
  ...
  ....
  =A(1)+2(1+2+.......+(n-1)+n)
  =2+2(n(n+1)/2)
  A(n+1)=2+n(n+1)
  there4 A(100)=2+100(99)=9902.
• 11 years agoHelpfull: Yes(2) No(1)