What are the next few numbers in the sequence? 1, 6, 18, 40, 75, 126, ...?

1,6,18,40,75,126..... is the given sequence

Now keep on keep on taking difference between the consecutive numbers and again & again for the obtained sequences

1,6,18,40,75,126...

5,12,22,35,51...

7,10,13,16...

3,3,3,3...

Stablised a relation in three steps so we can safely assume nth term as

A(n)=an^3+bn^2+cn^1+bn^0

putting one by one n=1,2,3,4 we get 4 eqn in four variables

a+b+c+d=1

8a+4b+2c+d=6

27a+9b+3c+d=18

64a+16b+4c+d=40

solve them to get a,b,c,d

hence we get nth term as

A(n)=(n^3)/2  +(n^2)/2  +n-1

By check we come to know that n-1 is not significant(i dont understand why,please add comment if anyone can)

so our new nth term is

A(n)=(n^3)/2  +(n^2)/2

and really it satisfies and returns all the values

using this we can get next any numbers of terms

196,288,549,600.....

Please inform if u find any mistake!!!!

Posted by Nishant Singh on 2007-04-09 12:08:00