Question

PLEASE HELLP 25 POINTS

Answer 1

Answer:

Step-by-step explanation:

You may need to clean things up a bit but suppose that

S(1) = a-1

S(2) = a^2 -1

Since this is a geometric series, the geometric ratio is given by

S(2)/ S(1) =  (a^2 -1)/ (a-1)

              =  (a+1)(a-1)/ (a-1)

              = a+1

Conclusion:

S(2) = (a+1) S(1) = (a+1) (a-1)

S(3)  = (a+1) S(2) = (a+1) (a+1) (a-1)   = (a+1)^ (3-1) (a-1)

S(4) = (a+1) S(3)  = (a+1) * (a+1)^2 (a-1) ) = (a+1)^(4-1) (a-1)

in general.....

S(n) = (a+1)^ (n-1) (a-1)

So

S(6) = (a+1)^ (6-1) (a-1)

     =  (a-1) (a+1) ^ 5

    =  a^6 + 4 a^5 + 5 a^4 - 5 a^2 - 4 a - 1

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Answer 2

Hello from MrBillDoesMath!

Answer:

a^6 + 4 a^5 + 5 a^4 - 5 a^2 - 4 a - 1

Discussion:

You may need to clean things up a bit but suppose that

S(1) = a-1

S(2) = a^2 -1

Since this is a geometric series, the geometric ratio is given by

S(2)/ S(1) =  (a^2 -1)/ (a-1)

               =  (a+1)(a-1)/ (a-1)

               = a+1

Conclusion:

S(2) = (a+1) S(1) = (a+1) (a-1)

S(3)  = (a+1) S(2) = (a+1) (a+1) (a-1)   = (a+1)^ (3-1) (a-1)

S(4) = (a+1) S(3)  = (a+1) * (a+1)^2 (a-1) ) = (a+1)^(4-1) (a-1)

in general.....

S(n) = (a+1)^ (n-1) (a-1)

So

S(6) = (a+1)^ (6-1) (a-1)

      =  (a-1) (a+1) ^ 5

     =  a^6 + 4 a^5 + 5 a^4 - 5 a^2 - 4 a - 1

Hope I didn't screw something here!

Thank you,

MrB