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# Algebra

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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?

Jun 3, 2024

#1
+1790
+1

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

\( a + b + c + d = 12\\ 8a + 4b + 2c + d = 44\\ 27a + 9b + 3c + d = 100\\ 64a + 16b + 4c + d = 186\\\)

Solving this, we get

\(a =1 \\ b= 6 \\ c= 7 \\ d= -2\)

This means we have a polynomial \(n^3 + 6n^2 + 7n - 2\)

Plugging in 5, we get 308.

Thanks! :)

Jun 3, 2024

#1
+1790
+1

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

\( a + b + c + d = 12\\ 8a + 4b + 2c + d = 44\\ 27a + 9b + 3c + d = 100\\ 64a + 16b + 4c + d = 186\\\)

Solving this, we get

\(a =1 \\ b= 6 \\ c= 7 \\ d= -2\)

This means we have a polynomial \(n^3 + 6n^2 + 7n - 2\)

Plugging in 5, we get 308.

Thanks! :)

NotThatSmart Jun 3, 2024
#2
+129829
0

I like your polynomial  approach  !!!!!

CPhill  Jun 3, 2024
#3
+1790
+1

Thanks! :)

NotThatSmart  Jun 3, 2024