+1533 Consecutive positive even integers are sorted into $100$ groups. Group $1$ includes $2,$ $4$ and $6.$ Group $2$ includes $8,$ $10,$ $12$ and $14.$ Group $3$ includes $16,$ $18,$ $20,$ $22$ and $24.$ Each successive group has one additional number in it than the previous group. What is the sum of the numbers in the $5$th group?
+1632 Nothing scary in listing out group 4 and 5, as they are pretty small: shouldn't take long :)
Group 4: 26, 28, 30, 32, 34, 36
Group 5: 38, 40, 42, 44, 46, 48, 50
Sum of Group 5 = 38 + 40 + 42 + 44 + 46 + 48 + 50, pair up the units digits to make a 10 for ease of calculation:
38 + 42, + 44 + 46, + 40 + 50, + 48
38 + 42 = 80
44 + 46 = 90
40 + 50 = 90
48 = 48
80 + 90 + 90, + 48 = 260 + 48 = 308
Or just use a calculator to get 308 lol (although not encouraged)
+129850 We can calculate a generating polynomial by the sum of differences
12 44 100 186 308
32 56 86 122
24 30 36
6 6
0
We have 3 rows of non-zero differences
So we can generate a 3rd power polynomial for the nth group sum
an^3 + bn^2 + cn + d where n is the nth group
We have this system
a + b + c + d = 12
8a + 4b + 2c + d = 44
27a + 9b + 3c + d = 100
64a + 16b + 4c + d = 186
Solving this gives us
a =1
b= 6
c= 7
d= -2
The generating polynomial for the sum of the nth group is :
n^3 + 6n^2 + 7n - 2
So.....the sum of the 5th group is
5^3 + 6*5^2 + 7*5 - 2 = 308
