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avatar+1537 

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?

 Feb 8, 2024
 #1
avatar+1632 
+1

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)

 Feb 8, 2024
 #2
avatar+129852 
+2

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

 

cool cool cool

 Feb 9, 2024

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