2nd difference → 5 7 9 11
3rd difference → 2 2 2
The 3rd difference is denoted by the term 6a.
Therefore in order to find a, I will replace the 3rd difference value
6a = 2
a = 2 = 1
6 3
By the given formula, Un = an3 + bn2 +cn + d, I will now be able to arrive to the generalized formula.
U1 = an3 + bn2 +cn + d
1 = 1 (1)3 + b(1)2 + c(1) + d
3
= b + c + d = 2 [equation 1]
3
U2 = 1 (2)3 + b(2)2 + c(2) + d
3
5 = 8+ 4b + 2c + d
3
= 4b + 2c + d = 7 [equation 2]
3
U3 = 1 (3)3 + b(3)2 + c(3) + d
3
14 = 9 + 9b + 3c + d
= 9b + 3c + d = 5 [equation 3]
[equation 2 – equation 1]
4b + 2c + d = 7
3
b + c + d = 2
3
= 3b + c = 5/3 [equation3]
- [equation 3 – equation 2]
9b + 3c + d = 5
4b + 2c + d = 7
3
= 5b + c = 8 [equation 5]
3
- [equation 5 – equation 4]
5b + c = 8
3
3b + c = 5
3
= 2b = 1
= b = ½
In order to find c I will replace b = ½ in equation 4
3b + c = 5
3
3*½+ c = 5
3
3+ c = 5
2 3
c= 10 – 9
6
c= 1
6
To find d I will now replace b and c in equation 1
½ + 1 + d = 2
6 3
4+ d = 2
6 3
d=2 - 2
3 3
d=0
Therefore
a = 1n3
3
b = 1n2
2
c = 1
6
Therefore I will now replace a, b and c to obtain the generalized formula which is
Un= 1n3+ 1n2 + 1 [formula 1]
3 2 6
I will be finding a formula with the extra number of squares added with the help of the sequence.
0 2 11 33 74 140
2 9 22 41 66
2nd difference →7 13 19 25
3rd difference → 6 6 6
The 3rd difference is denoted by the term 6a.
Therefore in order to find a, I will replace the 3rd difference value
6a = 6
Therefore a = 6= 1
6
a = 1
In order to find b c and d I will replace the value a = 1 in the given equation
Un = an3 + bn2 +cn + d
U1 = 1(1)3 + b (1)2 + c(1) +d
0 = 1 + b + c + d
b + c + d = -1 [equation 1]
U2 = 1 (2)3 + b(2)2 + c(2) + d
2 = 8 + 4b + 2c + d
=4b + 2c + d = -6 [equation 2]
U3 = 1(3)3 + b(3)2 + c (3) + d
11 = 27 + 9b + 3c + d
=9b + 3c + d = -16 [equation 3]
- [Equation 3 – equation 1]
4b + 2c + d = -6
b + c + d = -1
=3b + c = -5 [equation 4]
- [equation 3 – equation 2]
9b + 3c + d = -16
4b + 2c + d = -6
= 5b + c = -10 [equation 5]
- [equation 5 – equation 4]
5b + c = -10
3b + c = -5
= 2b = -5
= b = -5
2
In order to find c I will replace the value b = -5 in equation 5
2
5 * -5+ c = -10
2
-25 + c = -10
2
c = -10 + 25
2
c = -20 + 25
2
c = 5
2
In order to find d I will now replace b and c in equation 1
-5+ 5 + d = - 1
2 2
0 + d = - 1
Therefore
a=1
b= -5
2
c=5
2
d= -1
Therefore generalized formula is = n3 – 5n2 + 5n – 1 [formula 2]
2 2
After obtaining both the formulas, I will now be adding formula 1 and formula 2 to justify the formula I got in level 7.
1n3 + 1n2 + 1n
3 2 6
1n3 – 5n2 + 5n – 1
2 2
= 4n3– 2n2 + 8n – 1
3 3
Thus I do obtain the formula which I had obtained in level 7. This thus justifies that my formula is right.
