1,6,7,13,20 ...Find sum upto 52 terms.

• Sum of first 52 elements=352849027131.
  Observing the pattern of the given numbers, we conclude that T(n)=T(n-1)+T(n-2); where T(n) is the nth term of the given sequence of numbers. I used a computer program to generate few terms of this sequence and the sum till an arbitrary 'n'. The sums follow the pattern S(n)=T(n+2)-6; where S(n) is the sum of elements till n. Hence, by using the above pattern we can get sum of first 52 elements of the sequence. Here's the link to the program .
• 8 years agoHelpfull: Yes(5) No(1)
• Ans: 33
  Sol: 1+6=7
  6+7=13
  7+13=20
  13+20=33
  
• 8 years agoHelpfull: Yes(4) No(50)
• nth term =a+(n-1)d
  =1+(52-1)5
  256
• 8 years agoHelpfull: Yes(2) No(35)
• I thought its not in A.P and G.P also. so how we get the solution for this???>
• 8 years agoHelpfull: Yes(2) No(0)
• Its a fibonacci series...check the formula for nth term in google..it ll b like 1+√5/2
• 8 years agoHelpfull: Yes(2) No(1)
• 1,6,a+b,a+2b,a+3b,.................................,a+52b
  nth fib no=(n+2)th-2nd term=307
• 8 years agoHelpfull: Yes(1) No(7)
• ans by progrm
  
  int a = 1;
  int b = 6;
  int c;
  System.out.print(a +",");
  System.out.print(b +",");
  for(int n=0; n
• 8 years agoHelpfull: Yes(1) No(2)
• sn=n/2{2a+(n-1)d}
  =52/2{2*1(52-1)*5}
  = 6682
• 8 years agoHelpfull: Yes(1) No(4)
• ans)352849113670
• 8 years agoHelpfull: Yes(1) No(1)
• T(0)=1,T(1)=6
  T(N+2)=T(N)+T(N+1) WHERE N>0
  By using this we get the value
  T(1)...+T(52)=352849027131
  
  
  #include
  
  
  #include
  
  
  int main()
  
  
  {
  
  
  int func[100];
  
  
  func[1]=1;
  
  
  func[2]=6;
  
  
  int sum=7;
  
  
  for(int i=3;i
• 8 years agoHelpfull: Yes(0) No(6)
• The sum of the fibonacci number is[(n+2).Fibonacci No.-1]=[(54+2)54]-1=3023
• 8 years agoHelpfull: Yes(0) No(1)
• the answer is 368225351
  
  import java.io.*;
  
  class GFG {
  
  // Computes value of first
  // fibonacci numbers
  static int calculateSum(int n)
  {
  if (n
• 4 years agoHelpfull: Yes(0) No(0)