Question

calcule a soma das pg A) ( 4096, 1024,... 16) B) ( 3, -6, 12, .... 768 )

Answer 1

a)
16/4096 = (1/2)⁽ⁿ ⁻¹⁾
1/256 = (1/2)⁽ⁿ ⁻¹⁾
1/2⁸ = (1/2)⁽ⁿ ⁻¹⁾
2⁻⁸ = 2⁽⁻ⁿ ⁺¹⁾

-8 +1 = -n .(-1)
7 = n

s7 = 4096 [(1/2)^7 -1]/1/2 -1
s7 = -8192 .[1/128 -1]
s7 = -8192 . -127/128
s7 = 8128

b) 768/3 = (-2)⁽ⁿ ⁻¹⁾
256 = (-2)⁽ⁿ ⁻¹⁾
2⁸ = (-2)⁽ⁿ ⁻¹⁾
8 + 1 = n
9 = n

s9 = 3 . [(-2)^9 -1]/-2 -1
s9 = -1 . -511
s9 = 511