Question

How do you find the sum of the finite geometric sequence of `sum_(j=1)^6 32(1/4)^(j-1)` ?

Answer 1

Use a formula found in this reference

`sum_(j=1)^n ar^(j-1)=a(1-r^n)/(1-r); r!=1`

Explanation:

Substituting into the formula, `n=6, a = 32 and r = 1/4`:

`sum_(j=1)^6 32(1/4)^(j-1)=32(1-(1/4)^6)/(1-1/4)`

`sum_(j=1)^6 32(1/4)^(j-1)=32(1-1/4096)/(1-1/4)`

`sum_(j=1)^6 32(1/4)^(j-1)=1365/32`