+1517 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?
+1950 Although I don't really recommend it, we could just list out all the numbers since groups 5 and 4 are really not that big.
We have
Group 4 - 26, 28, 30, 32, 34, 36
Group 5 - 38, 40, 42, 44, 46, 48, 50
Adding the numbers in group up, we get 308, which is our answer.
We could also generate a polynomial equation to help us.
We could put it in the form of a polynomial with 3rd degree in ax^3+bx^2+cx+d
From the problems above, we can get the system
a+b+c+d=128a+4b+2c+d=4427a+9b+3c+d=10064a+16b+4c+d=186
Solving this, we get
a=1b=6c=7d=−2
This means we have a polynomial n3+6n2+7n−2.
Plugging in 5, we get 308.
Thanks! :)
+1950 Although I don't really recommend it, we could just list out all the numbers since groups 5 and 4 are really not that big.
We have
Group 4 - 26, 28, 30, 32, 34, 36
Group 5 - 38, 40, 42, 44, 46, 48, 50
Adding the numbers in group up, we get 308, which is our answer.
We could also generate a polynomial equation to help us.
We could put it in the form of a polynomial with 3rd degree in ax^3+bx^2+cx+d
From the problems above, we can get the system
a+b+c+d=128a+4b+2c+d=4427a+9b+3c+d=10064a+16b+4c+d=186
Solving this, we get
a=1b=6c=7d=−2
This means we have a polynomial n3+6n2+7n−2.
Plugging in 5, we get 308.
Thanks! :)
