January 31st, 2015, 09:33 AM  #1 
Newbie Joined: Nov 2014 From: Cottingham Posts: 8 Thanks: 0 Math Focus: Surds!  Nth term! How?
The sequence is: 1, 3, 6, 10, 15, 21, 28 So the interval is: 2, 3, 4, 5, 6, 7 And that interval is: 1 Soo, the nth term may be n^2/2 + n/2 But that seems waaaay to complicated, is there an easier way to, say, work out the 100th term (which was the question) 
January 31st, 2015, 09:42 AM  #2 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
This sequence of numbers is famous , they are called triangular numbers. You don't have to square anything , one of the numbers , n or (n + 1) will be even so divide it by 2 and multiply the result with the other. Just plug and play. 
January 31st, 2015, 09:47 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,634 Thanks: 2620 Math Focus: Mainly analysis and algebra 
More explicitly $${n^2 \over 2}+{n \over 2} = \tfrac12(n^2 +n) = \tfrac12n(n+1)$$

January 31st, 2015, 11:26 AM  #4  
Senior Member Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176  Quote:
\begin{aligned}a_1=1\\ a_2=1+2=3 \\ a_3=1+2+3=6\\ a_4=1+2+3+4=10 \\ .\\.\\. \\ a_n=1+2+3+ ... +n =\dfrac{n(n+1)}{2} \end{aligned} \\\;\\ \\\;\\ a_{100}=\dfrac{100\cdot101}{2}=5050$  
January 31st, 2015, 11:49 AM  #5 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
Or, search the oeis; 1,3,6,10,15,21,28  OEIS 
February 1st, 2015, 09:40 AM  #6  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
1 = 1 . . . first term 1 + 2 = 3 . . . second term 1 + 2 + 3 = 6 . . . third term 1 + 2 + 3 + 4 = 10 . . . fourth term . . . Sum = 1 + 2 + 3 + . . . + 98 + 99 + 100 . . . hundredth term Sum = 100 +99 + 98 + ... + 3 + 2 + 1 . . . . . hundredth term Twice the sum = 101 + 101 + 101 + ... + 101 + 101 + 101 There are 100 columns. 2*Sum = 100*101 Sum = (100*101)/2 You can finish it.  

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