1,3,6,10,15,21,28........................so on up to 300 terms then find the sum

• 1 ,3, 6, 10 is of the form n*(n+1)/2
  n = 300
  nth term = 300*(301)/2 = 45150
  Then sum of these terms 1+3+6+10+15+......45150 ll be
  Sum of n*(n+1)/2 terms
  Sum of n*(n+1)/2 terms = 1/2* sum of [n^2 + n] terms
  Sum of n^2 terms => n(n+1)(2n+1)/6
  sum of n terms => n(n+1)/2
  So Sum ll be 1/2 * [n(n+1)(2n+1)/6 + n(n+1)/2]
  n(n+1)(2n+4)/12
  n = 300
  So 300(301)(604)/12 = 4545100
  
  Ans : 4545100
  
  
• 10 years agoHelpfull: Yes(38) No(0)
• each term in the series follows the equation-n(n+1)/2, where n is the position
  therefore for the sum of 300 terms of the series , we will apply summation over n(n+1)/2,
  hence it will expand to-- ((n(n+1)(2n+1))/6)+(n(n+1)/2)
  which will compute to 300*301*604/12
  =4545100
• 10 years agoHelpfull: Yes(8) No(1)
• ans==4545100
• 10 years agoHelpfull: Yes(3) No(3)
• sn=n(a+l)/2.....
  n=300,a=1,l=300
  sn=300(301)/2
  ans=45150
• 10 years agoHelpfull: Yes(3) No(4)
• =1+3+6+10+15+21+28+36+.............................up to 300 terms
  =4+16+36+64+..................................up to 150 terms
  =2^2+4^2+6^2+8^2+...........................up to 150 terms
  =2^2*(1^2+2^2+3^2+4^2...........................up to 150 terms)
  =4*(n*(n+1)*(2*n+1)/6)
  put n=150
  =2*(150*151*301)/3
  =4545100
  
• 9 years agoHelpfull: Yes(3) No(0)
• n=300 and a=1 and d=1 so now we no the formula of ap series n/2[2a+(n-1)d]
  300/2[2*1+(300-1)*1]
  45150
• 10 years agoHelpfull: Yes(0) No(5)
• can someone tell me wat will be d diagram..i am got getting the question properly
  
• 10 years agoHelpfull: Yes(0) No(0)
• n=300,a=1,d=1,(the diffrence between successive terms in A.P..1,3-1,6-3,10-6,15-15....) we know the formula.sum of the nth terms in A.P=n/2(a+l)
  =300/2(1+300)=45,150
• 10 years agoHelpfull: Yes(0) No(2)
• term are in the of n*(n+1)/2
  sum of n^2 is n*(n+1)*(2*n+1)/12
  sum of n term is n*(n+1)/2
  so, sn=(n*(n+1)*(2*n+1)/12+n*(n+1)/2)/2
  so ans 4545100
  
• 9 years agoHelpfull: Yes(0) No(0)