Question

How do you find the sum of the first 40 terms of the geometric sequence 2, 8, 32, 128?

Answer 1

The questions asked "how do you find.." In other words: it is not asking for the actual value!
`=>s=(2(4^40-1))/(4-1)`

Explanation:

I spot that:
`2xx4=8`
`8xx4=32`
`32xx4=128`

Also that
`2xx4^0=2`
`2xx4^1=8`
`2xx4^2=32`
`2xx4^3=128`

So this sequence is:`" "2xx4^(n-1)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~
Also
`1/2xx4^1=2`
`1/2xx4^2=8`
`1/2xx4^3=32`
`1/2xx4^4=128`

So this sequence is: `" "1/2xx4^n`
'~~~~~~~~~~~~~~~~~~~~~~~~
=>`1/2xxn^4=2xx4^(n-1)`
Multiply bot sides by 2
`=>2/2xxn^4=4xxn^3`
`=>n^4=n^4->" both are correct forms of the same thing!"`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Solving the question")`

`color(brown)("Considering a general case")`

Let the sum of a sequence be `s`
Let the sum be:`" "s=color(magenta)(ar^1)+ar^2+ar^3+...+ar^n`

so `" "sr= " "ar^2+ar^3+...+ar^ncolor(blue)(+ar^(n+1))`

`color(white)(.)`

Thus`" "sr -s= color(blue)(ar^(n+1))color(magenta)(-ar^1)`

`s(r-1)=ar(r^n-1)`

`s=(ar(r^n-1))/(r-1)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let
`a=1/2`
`r=4`
`n=40`

`=>s=(2(4^40-1))/(4-1)`