a) b1 = 2
bn+1 = 3bn
=> bn+1 / bn = 3 => q = 3
=> bn = b1×q^(n-1) => bn = 2×3^(n-1)
d) b1 = 10
bn+1 = 1/5 bn => bn+1 / bn = 1/5 => q = 1/5
=> bn = b1×q^(n-1) => bn = 10×(1/5)^(n-1) => bn = 10×5^(1-n) =>
=> bn = 2×5×5^(1-n) => bn = 2×5^(2-n)
Si asa se fac toate.