1,6,7,13,20 33 so on......find the sum of first 52 terms

• as 1+6+7 is sum upto 3rd tem n it is 5th term -6, similarly
  as 1+6+7+13+20+33+53 is sum upto 7th term and is equal to 9th term 139-6=133
  hence sum upto 52nd term is 54th term-6
  using golden ratio we can solve this, bt to be more accurate we calculate a few more terms. 1,6,7,13,20,33,53,86,139,225,364,589,953,1542....
  thus 54th term= 1542*(1.6180339)^40
  hence sum upto 52nd term is 352849113676-6=352849113670
• 10 years agoHelpfull: Yes(6) No(1)
• by mistake i have given the solution here for the below question ,plz excuse
• 10 years agoHelpfull: Yes(2) No(6)
• # include
  # include
  int main()
  {
  long int a = -4, b = 5, c = 0, i, n, sum = 0 ;
  
  printf("Enter the limit : ") ;
  scanf("%d", &n) ;
  printf("nThe fibonacci series is : nn") ;
  for(i = 1 ; i
• 10 years agoHelpfull: Yes(1) No(26)
• long int a = -4, b = 5, c = 0, i, n, sum = 0 ;
  printf("Enter the limit : ") ;
  scanf("%d", &n) ;
  printf("nThe fibonacci series is : nn") ;
  for(i = 1 ; i
• 10 years agoHelpfull: Yes(1) No(22)
• 352849027131.
• 10 years agoHelpfull: Yes(1) No(0)
• ths is not c program compitition @suman
• 10 years agoHelpfull: Yes(1) No(0)
• MATHEMATICALLY PROCESS IS TOO LENGTHY
  SO........EASIEST WAY.......after running the program enter value of n=52
  # include
  # include
  int main()
  {
  long int a = -4, b = 5, c = 0, i, n, sum = 0 ;
  
  printf("Enter the limit : ") ;
  scanf("%d", &n) ;
  printf("nThe fibonacci series is : nn") ;
  for(i = 1 ; i
• 10 years agoHelpfull: Yes(0) No(41)
• for(i=1;i
• 10 years agoHelpfull: Yes(0) No(20)
• 2(10C3*3!*7!+ 10C4*4!*6!+ 10C5*5!*5!+10C6*6!*4!+ 10C7*7!*3!+10C8*8!*2!+ 10C9*9!*1!+ 10C10*10!)
• 10 years agoHelpfull: Yes(0) No(11)
• as of previous formula here also is S(n)= S(n+2)-S(2)
  but 54th term is unknown..
• 10 years agoHelpfull: Yes(0) No(4)
• Sn=n(a1+an)/2
  Here,a1=1 and an=(1+(52-1)d) where, d=(6-1)
  hence,an=307
  Sn=52(1+307)/2
  so...8008 (ans)
• 10 years agoHelpfull: Yes(0) No(15)