Find the sum of n terms of the series
1+2+4+7+11+16++22+...(nth term)
First find the successive sums of the 7 terms given:
1 = 1 = S(1)
1 + 2 = 3 = S(2)
1 + 2 + 4 = 7 = S(3)
1 + 2 + 4 + 7 = 14 = S(4)
1 + 2 + 4 + 7 + 11 = 25 = S(5)
1 + 2 + 4 + 7 + 11 + 16 = 41 = S(6)
1 + 2 + 4 + 7 + 11 + 16 + 22 = 63 = S(7)
Write those in a list beside the 1st 7 positive
integers:
n S(n)
1 1
2 3
3 7
4 14
5 25
6 41
7 63
Form the column of 1st differences by
writing beside each number the difference
between it and the number just below it
n S(n) d1
1 1 2
2 3 4
3 7 7
4 14 11
5 25 16
6 41 22
7 63
Form the column of 2nd differences by
writing beside each number the difference between
it and the number just below it
n S(n) d1 d2
1 1 2 2
2 3 4 3
3 7 7 4
4 14 11 5
5 25 16 6
6 41 22
7 63
Form the column of 3rd differences the
same way.
n S(n) d1 d2 d3
1 1 2 2 1
2 3 4 3 1
3 7 7 4 1
4 14 11 5 1
5 25 16 6
6 41 22
7 63
Since the column of 3rd differences is
constant, we assume a 3rd degree equation
for S(n):
S(n) = An² + Bn² + Cn + D
Since we have only 4 unknowns, we only need
the first four sums:
S(1) = 1 = A*1³ + B*1² + C*1² + D
S(2) = 3 = A*2³ + B*2² + C*2² + D
S(3) = 7 = A*3³ + B*3² + C*3² + D
S(4) = 14 = A*4³ + B*4² + C*4² + D
Simplifying, we have this system of
4 equations in 4 unknowns:
A + B + C + D = 1
8A + 4B + 2C + D = 3
27A + 9B + 3C + D = 7
64A + 16B + 4C + D = 14
We solve this system of equations and get
A = 1/6, B = 0, C = 5/6, D = 0
So the equation
S(n) = An² + Bn² + Cn + D
becomes
S(n) = (1/6)n³ + (5/6)n
or simplified in factored form:
S(n) = n(n²+5)/6
Edwin
