Find the 100th term of the following series - 2,4,8,14,22,32...

• t(n+1)=n^2+n+2
  t(0+1)=0+0+2=2
  t(1+1)=1^2+1+2
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  t(99+1)=99^2+99+2=9902
• 11 years agoHelpfull: Yes(38) No(2)
• 2nd term -1st term=1*2
  3rd term -2nd term=2*2
  4th term -3rd term=3*2
  .....................
  .....................
  nth term -(n-1)th term=(n-1)*2
  ------------------------------
  sum:nth term -1st term=[1+2+3....(n-1)]*2(as rests are cancelld)
  so 100th term=(100*99/2)*2+2(as sum=n(n-1)/2,n=100,1st term =2)
  =9902ans
• 11 years agoHelpfull: Yes(5) No(1)
• ANS IS : 9902
  a(n) = 2 + n(n - 1)
  a(100)=2+100(100-1)
  =9902
• 9 years agoHelpfull: Yes(3) No(0)
• t(n)=t(n-1)+2n ;n>0
  t(0)=2
  t(1)-t(0)=2
  t(2)-t(1)=4
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  t(100)-t(99)=200
  summing up we get t(100)=(2*100*101/2)+t(0)=10102
• 11 years agoHelpfull: Yes(2) No(7)
• 2 4 8 14 22 32
  diff 2 4 6 8 10
  again diff 2 2 2 2
  now apply formula tn=a+(n-1)d
  t100=2+(99)2
  t100=200
• 11 years agoHelpfull: Yes(2) No(6)
• @sumit gupta----
  actually the t(n) comes from the given logi....its not aaplicable for all questions..
  
  s(n)=3+6+12+...+t(n)------eq 1
  s(n)= 3+6+12+...+t(n)+t(n+1)------eq2
  eq1-eq2
  t(n+1)=3+(3*n)/2+(3*n*n)/2
  this is d correct way....enjoy
• 11 years agoHelpfull: Yes(1) No(1)
• @rupesh
  Find the 100th term of the following series - 3,6,12,21,33,48...
  now what will be the logic
• 11 years agoHelpfull: Yes(0) No(0)
• can we solve the series 3,6,12,21,33.... using the above method i.e.
  t(n+1) = n^2+ n+2?
• 11 years agoHelpfull: Yes(0) No(1)
• @sumit...u rfinding the nth term for series of common differences and not for the given series!!
• 11 years agoHelpfull: Yes(0) No(0)
• .Now, How can you find any term like say 22 if you're given 2 only and this difference. Just the some of the differences plus 2.=> 2+ (2+4+6+8) = 22. Hence, 100th term = Sum of AP (2,4,6,8...) till 99 terms + 2=> [ 99*(2+98*2)/2 ] + 2 = 99*99 + 2 = 9801 + 2 = 9803.
• 11 years agoHelpfull: Yes(0) No(7)
• here nth term =2*n-1+n-1th term
  so 100 term=2*99+99th term
  99th term=2*98+98th term and so on
  s0
  2(99+98.....1)+2=9704
• 11 years agoHelpfull: Yes(0) No(3)