`sum_(i = 1)^7 64(-1/2)^(i - 1)` Find the sum of the finite geometric series.

`sum_(i=1)^7 64 (-1/2)^(i-1)`

The given summation notation has a form

`sum_(i=1)^i a_1 (r)^(i-1)`

Base on this, the first term and common ratio of the geometric series can be identified.  The values are a1=64 and r=-1/2.  

Plugging in the values of the a1 and r to the formula of geometric series

`S_n = a_1*(1-r^n)/(1-r)`

the sum of the first seven terms will be:

`S_7 = 64 *(1-(-1/2)^7)/(1-(-1/2))`

`S_7=64*((1-(-1/128))/(1-(-1/2))`

`S_7=64(1+1/128)/(1+1/2)`

`S_7=64*43/64`

`S_7=43`

Therefore,   `sum_(i=1)^7 64(-1/2)^(i-1)=43`...

`sum_(i=1)^7 64 (-1/2)^(i-1)`

The given summation notation has a form

`sum_(i=1)^i a_1 (r)^(i-1)`

Base on this, the first term and common ratio of the geometric series can be identified.  The values are a1=64 and r=-1/2.  

Plugging in the values of the a1 and r to the formula of geometric series

`S_n = a_1*(1-r^n)/(1-r)`

the sum of the first seven terms will be:

`S_7 = 64 *(1-(-1/2)^7)/(1-(-1/2))`

`S_7=64*((1-(-1/128))/(1-(-1/2))`

`S_7=64(1+1/128)/(1+1/2)`

`S_7=64*43/64`

`S_7=43`

Therefore,   `sum_(i=1)^7 64(-1/2)^(i-1)=43` .