find the nth term of the series 1,3,6,11,19,31,48.... Hence duduce a formula for calculating the sum of the series

Question

plz help me show working

Answer 1

Ok, revising my difference calculations

1
3 2
6 3 1
11 5 2 1
19 8 3 1
31 12 4 1
48 17 5 1

ahhh, so the third difference is constant, so we have a cubic
let the term = an^3 + bn^2 + cn + d
when n = 1:
1 = a + b + c + d
when n = 2
3 = 8a + 4b + 2c + d
when n = 3
6 = 27a + 9b + 3c + d
when n = 4
11 = 64a + 16b + 4c + d

solving 4 equations in 4 variables is rather tedious, and I will assume you know how to do that
I let Wolfram do it for me:

http://www.wolframalpha.com/input/?i=pattern+1,3,6,11,19,31,48

term(n) = (1/6)(n^3 - 3n^2 + 14n - 6)

now some more fun begins:
Sum(1) = 1
sum(2) = 1 + 3 = 4
sum(3) = 4 + 6 = 10
sum(4) = 10 + 11 = 21
sum(5) = 21 + 19 = 40
sum(6) = 40 + 31 = 71
sum(7) = 71 + 48 = 119

again using Wolfram to find a pattern
http://www.wolframalpha.com/input/?i=pattern+1,4,10,21,40,71,119

Term(n) = (1/24)(n^4 - 2n^3 + 23n^2 + 2n)

WoW!!! a quartic! Hope this wasn't a test question.

btw, I am a guy, not a miss

Answer 2

thanks