Please help me solve this problem?!?!
Find the first six terms and the sixth partial sum of the sequence whose nth term is
an = 5n2 − n.
a1 =
a2 =
a3 =
a4 =
a5 =
a6 =
S6 =
well, finding the first six terms is easy, right?
An = 5n^2-1, so here's a table:
n An
1 5*1^2-1 = 5-1 = 4
2 5*2^2-2 = 20-2 = 18
3 5*3^2-3 = 45-3 = 42
4 5*4^2-4 = 80-4 = 76
5 5*5^2-5 = 125-5 = 120
6 5*6^2-6 = 180-6 = 174
Now just add 'em up to get S6
To find the first six terms of the sequence, we substitute the values of n into the formula for the nth term, an = 5n^2 - n:
a1 = 5(1)^2 - 1
= 5(1) - 1
= 5 - 1
= 4
a2 = 5(2)^2 - 2
= 5(4) - 2
= 20 - 2
= 18
a3 = 5(3)^2 - 3
= 5(9) - 3
= 45 - 3
= 42
a4 = 5(4)^2 - 4
= 5(16) - 4
= 80 - 4
= 76
a5 = 5(5)^2 - 5
= 5(25) - 5
= 125 - 5
= 120
a6 = 5(6)^2 - 6
= 5(36) - 6
= 180 - 6
= 174
Therefore, the first six terms of the sequence are:
a1 = 4
a2 = 18
a3 = 42
a4 = 76
a5 = 120
a6 = 174
To find the sixth partial sum, we add up the first six terms:
S6 = a1 + a2 + a3 + a4 + a5 + a6
= 4 + 18 + 42 + 76 + 120 + 174
= 434
Therefore, the sixth partial sum of the sequence is 434.
To find the first six terms of the sequence whose nth term is given by an = 5n^2 - n, we can plug in the values of n from 1 to 6 in the formula.
a1 = 5(1^2) - 1 = 5(1) - 1 = 5 - 1 = 4
a2 = 5(2^2) - 2 = 5(4) - 2 = 20 - 2 = 18
a3 = 5(3^2) - 3 = 5(9) - 3 = 45 - 3 = 42
a4 = 5(4^2) - 4 = 5(16) - 4 = 80 - 4 = 76
a5 = 5(5^2) - 5 = 5(25) - 5 = 125 - 5 = 120
a6 = 5(6^2) - 6 = 5(36) - 6 = 180 - 6 = 174
Now, to find the sixth partial sum, S6, we need to add up the first six terms:
S6 = a1 + a2 + a3 + a4 + a5 + a6
= 4 + 18 + 42 + 76 + 120 + 174
= 434
Therefore, the first six terms of the sequence are:
a1 = 4, a2 = 18, a3 = 42, a4 = 76, a5 = 120, a6 = 174. And the sixth partial sum, S6, is 434.