1,6,7,13,20,33…………………………and so on find the sim of 52th term

• Ans: 352849113670
  1st and 2nd terms are 1,6
  3rd term is 7 which is sum of first+second.
  4th term is 13 which is sum of second + third.
  Expanding the series further by adding a few more terms, it is
  1, 6, 7, 13, 20, 33, 53, 86, 139 and so on.
  
  Now if we add the terms for upto 1,2,3,4,5,6 terms it is
  1, 7, 14, 27, 47, 80, 133, 219 and so on.
  
  Hence we can infer the sum upto Nth term is (N+2)nd term - 6
  As 1+6+7 is sum upto 3 terms and it is 5th term minus 6= 20-6 = 14
  As 1+6+7+13+20+33+53 is sum upto 7 terms and it is 9th term minus 6 = 139-6 = 133
  
  Hence sum upto 52nd term is nothing but 54th term -6.
  
  Now our task is to find 54th term.
  Using the golden ratio we can easily find the 54th term. But we will just find few more terms for accuracy.
  1, 6, 7, 13, 20, 33, 53, 86, 139, 225, 364, 589, 953, 1542 which is upto 14 terms.
  
  So 54th term = 1542 * (1.6180339)^40 = 352849113676 approx
  Hence sum will be 352849113676-6 = 352849113670
• 9 years agoHelpfull: Yes(3) No(6)
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• can anyone tell the correct answer
  
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• 32951280099
  
  http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibtable.html
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• Sorry frnds its wrong Answer...
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• 1+6=7
  6+7=13
  7+13=20
  13+20=33
  20+33=53
  
• 9 years agoHelpfull: Yes(0) No(5)