Final answer:
A. The series 4 + 16 + 64 + 256 + ... can be written in sigma notation as Σ(4^(n-1)). B. The series 3/4 + 4/5 + 5/6 + ... can be written in sigma notation as Σ((n+2)/(n+3)). C. The series 49m⁶ + 64m² + 81m⁸+ 100m⁹+ 121m¹⁰ can be written in sigma notation as Σ((n+6)^2 * m^(2n)).
Step-by-step explanation:
A. The given series 4 + 16 + 64 + 256 + ... can be written in sigma notation as Σ(4^(n-1)), where n ranges from 1 to infinity. This means that the first term of the series is 4^(1-1) = 4^0 = 1, the second term is 4^(2-1) = 4^1 = 4, the third term is 4^(3-1) = 4^2 = 16, and so on.
B. The given series 3/4 + 4/5 + 5/6 + ... can be written in sigma notation as Σ((n+2)/(n+3)), where n ranges from 0 to infinity. This means that the first term of the series is (0+2)/(0+3) = 2/3, the second term is (1+2)/(1+3) = 3/4, the third term is (2+2)/(2+3) = 4/5, and so on.
C. The given series 49m⁶ + 64m² + 81m⁸+ 100m⁹+ 121m¹⁰ can be written in sigma notation as Σ((n+6)^2 * m^(2n)), where n ranges from 0 to infinity. This means that the first term of the series is (0+6)^2 * m^(2*0) = 6^2 * m^0 = 36 * 1 = 36, the second term is (1+6)^2 * m^(2*1) = 7^2 * m^2 = 49m^2, the third term is (2+6)^2 * m^(2*2) = 8^2 * m^4 = 64m^4, and so on.