SOLUTION: 1) Evaluate the series 5 Σ (3)(1/2)^k k=1 A. 93/32 B. 45/16 C. 23/8 D. 11/4 2) Evaluate the partial sum 20 Σ (26+4k) k=5 A. 986 B. 1060 C. 1140 D. 1

Algebra ->  Sequences-and-series -> SOLUTION: 1) Evaluate the series 5 Σ (3)(1/2)^k k=1 A. 93/32 B. 45/16 C. 23/8 D. 11/4 2) Evaluate the partial sum 20 Σ (26+4k) k=5 A. 986 B. 1060 C. 1140 D. 1      Log On


   



Question 1085497: 1) Evaluate the series
5
Σ (3)(1/2)^k
k=1
A. 93/32
B. 45/16
C. 23/8
D. 11/4
2) Evaluate the partial sum
20
Σ (26+4k)
k=5
A. 986
B. 1060
C. 1140
D. 1216
3) Find the partial sum of the first 8 terms​of the​ series
8+16+32+...
A. 1024
B. 2040
C. 2048
D. 2464
4) Evaluate the partial sum
9+12+15+...+51+54+57
A. 504
B. 561
C. 621
D. 698

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
k=1, product is 3/2
2--3/4
3--3/8
4--3/16
5--3/32
common denominator is 32, numerator is 93, 93/32,
A
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at 5, it is 46. At 20, it is 106
at 12, it is 74, at 13, it is 78. The average term is 76. There are 16 terms, average is 76, total 1216.
D
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a1=8; an=a1*2^(n-1)
for 8th term is is a1(1-2^8)/(1-2)=8*(-255/-1)=8*255=2040
8+16+32+64+128+256+512+1024=2040
C
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Partial sum between 9 and 57, with d=3
There are 17 terms, there is a middle term (the ninth, which is a1+(3*8)=33) so 17 times the average, which is 33=561.
B