The sums can be calculated as follows:
1) The first sum:
\sum_(k=0)^5 50((1)/(8))^(k)
= 50(1/8)^0 + 50(1/8)^1 + 50(1/8)^2 + 50(1/8)^3 + 50(1/8)^4 + 50(1/8)^5
= 50 + 6.25 + 0.78125 + 0.09765625 + 0.01220703125 + 0.00152587890625
≈ 57.1674805
2) The second sum:
\sum_(k=0)^6 27((1)/(3))^(k)
= 27(1/3)^0 + 27(1/3)^1 + 27(1/3)^2 + 27(1/3)^3 + 27(1/3)^4 + 27(1/3)^5 + 27(1/3)^6
= 27 + 9 + 3 + 1 + 1/3 + 1/9 + 1/27
= 36
3) The third sum:
\sum_(k=0)^6 33((1)/(2))^(k)
= 33(1/2)^0 + 33(1/2)^1 + 33(1/2)^2 + 33(1/2)^3 + 33(1/2)^4 + 33(1/2)^5 + 33(1/2)^6
= 33 + 16.5 + 8.25 + 4.125 + 2.0625 + 1.03125 + 0.515625
≈ 65.484375
Therefore, the sums are approximately:
57.1674805, 36, and 65.484375.
what are the sums? \sum_(k=0)^5 50((1)/(8))^(k) \sum_(k=0)^6 27((1)/(3))^(k) \sum_(k=0)^6 33((1)/(2))^(k)
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