Series

The series is 0,0,1,1,2,3,5,6,8,10,13,15,18,21,25,28,32,36,..............

Find the nth term of the series.

 n = series-I series-II series-III series-IV n is even, divisible by 2 odd, divisible by 4 after subtracting 3 even, divisible by 4 odd, divisible by 4 after subtracting 5 x-value (x2) (x3) (x4) (x5) part sum Value nth term ->

( not showing result for 1st,2nd and 3ed term. ) link: www.mathaddict.net/token3.htm

The nth term consists of 4 series---

Series 1: add 2+add 6 +........... = Math.Floor(2/4) + Math.floor(6/4)+...... up to x2 value of series-I

Series II: add 3 +add 7 + ......=(3+1)/4 +(7+1)/4 +......=([6+(x3-1)4](x3/2) +x3) /4=x3(1+x3)/2

Series III: add 4+ add 8+.....=((4/4)-1)) + ((8/4)-1)) +....= ([8+(x4-1)4]x4/8) - x4 = x4(x4-1)/2

Series IV: add 5 +add 9 +....=(5-1)/4 +(9-1)/4 +....=[( [10+(x5-1)4](x5/2)) - x5 ] / 4

This series becomes useful when we calculate the arrangements of  3 tokens taken out from n number of tokens starting from 1 and putting them 1 each in 3 identical boxes and trying to find out for what aggregate sum of 3 tokens,

the no. of arrangements is maximum. Example is we have 8 tokens staring from no. 1,2,3.......8. They are drawn out and put in 3 identical boxes with 1 token in each box. The sum of token numbers can be any integer in between 6 (3+2+1)

to 21(8+7+6). We have to find out for which sum , the no. of arrangements is maximum. We have separately found out that for average sum i.e. (6+21)/2= 13.5 i.e. 13,14-> the no. of alternate configurations is maximum. What is the

maximum configuration. From the above series, we can find out. In this case we have 8 tokens with no. 1,2,3,4,5,6,7,8. Hence n=8. The nth term i.e. 8th term will indicate the amount of maximum configuration possible which is 6. For

aggregate sum of 13 -(3,4,6),(2,5,6),(2,4,7),(1,5,7),(1,4,8),(2,3,8) are the configurations.