precalculus
posted by Ama .
Find the first six partial sums S1, S2, S3, S4, S5, S6 of the sequence.
2, 5, 8, 11, . . .

The sum of the members of a arithmetic progression :
Sn = ( n / 2 ) * [ 2 a1 + ( n  1 ) * d ]
d = common difference of successive members of arithmetic progression
In this case :
S1 = ( 1 / 2 ) * [ 2 * 2 + ( 1  1 ) * 3 ]
S1 = ( 1 / 2 ) * (4 + 0 * 3 )
S1 = ( 1 / 2 ) * 4 = 2
S2 = ( 2 / 2 ) * [ 2 * 2 + ( 2  1 ) * 3 ]
S2 = 1 * ( 4 + 1 * 3 )
S2 = 4 + 3 = 7
S3 = ( 3 / 2 ) * [ 2 * 2 + ( 3  1 ) * 3 ]
S3 = ( 3 / 2 ) * ( 4 + 2 * 3 )
S3 = ( 3 / 2 ) * ( 4 + 6 )
S3 = ( 3 / 2 ) * 10
S3 = 30 / 2 = 15
S4 = ( 4 / 2 ) * [ 2 * 2 + ( 4  1 ) * 3 ]
S4 = 2 * ( 4 + 3 * 3 )
S4 = 2 * ( 4 + 9 )
S4 = 2 * 13
S4 = 26
S5 = ( 5 / 2 ) * [ 2 * 2 + ( 5  1 ) * 3 ]
S5 = ( 5 / 2 ) * ( 4 + 4 * 3 )
S5 = ( 5 / 2 ) * ( 4 + 12 )
S5 = ( 5 / 2 ) * 16
S5 = 80 / 2 = 40
S6 = ( 6 / 2 ) * [ 2 * 2 + ( 6  1 ) * 3 ]
S6 = 3 * ( 4 + 5 * 3 )
S6 = 3 * ( 4 + 15 )
S6 = 3 * 19
S6 = 57