Question

Compute each sum below. If applicable, write your answer as a fraction.
`2 + 2( ( - 1)/(4) ) + 2( ( - 1)/(4)) {}^(2) + ... + 2( ( - 1)/(4) ) {}^(6)`

Answer 1

The Solution.

The series is a geometric series since the condition below holds:

`\begin{gathered} r=(T_2)/(T_1)=(T_3)/(T_2) \\ \text{Where T}_1=first\text{ term=2} \\ T_2=\sec ond\text{ term = 2(}(-1)/(4)) \\ \\ T_3=third\text{ term = 2(}(-1)/(4))^2 \end{gathered}`

So, the common ratio (r) is:

`r=(T_2)/(T_1)=(2(-(1)/(4)))/(2)=-(1)/(4)`

The sum of the series is given as below:

`\begin{gathered} S_n=(a(1-r^n))/(1-r)\text{ , r}<1 \\ \text{Where s}_n=\text{ s um of n terms of the series} \\ a\text{ =first term = 2} \\ r\text{ =common ratio = -}(1)/(4) \\ n=\text{ number of terms = 6} \end{gathered}`

Substituting the above values into the stated formula for sum, we get

`S_6=\frac{2\lbrack(1-(-(1)/(4))^6\rbrack^{}}{1-(-(1)/(4))}`
`\begin{gathered} S_6=(2(1-(1)/(4096)))/(1+(1)/(4)) \\ \\ S_6=(2*(4095)/(4096))/((5)/(4)) \end{gathered}`