Question

question from chapter 2 sequences and series​

Answer 1

Answer:

see below

Step-by-step explanation:

€an = 2(3^n - 1) ; an = ?

an = €an - €a(n-1)

but €a(n-1) = 2[3^(n-1) - 1]

now...

an = 2(3^n - 1) - 2[3^(n-1) - 1]

an = 2 [ 3^n - 1 - 3^(n-1) + 1]

an = 2 [ 3^n - 3^n /3]

an = 2 3^n (2/3)

an = 4 3^(n-1)

hence proved

Note

€ : sigma (summation)