Given sum of terms Sn = 4^(n+1)-1, demonstrate that is sum of terms of geometric progresion?

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You need to evaluate the formula of general term of geometric sequence, hence, you should use the following formula, such that:

`a_n = S_n - S_(n-1)`

`a_n = 4^(n+1) - 1 - 4^(n-1+1) + 1`

Reducing duplicate members yields:

`a_n = 4^(n+1)- 4^n`

You may prove that the given summation represents the summation of n terms of a geometric sequence if you test the validity of the following formula, such that:

`(a_n)^2 = a_(n-1)*a_(n+1)`

Replacing `4^(n-1+1)- 4^(n-1)` for `a_(n-1)` and `4^(n+1+1)- 4^(n+1)` for `a_(n+1)` yields:

`(4^(n+1)- 4^n)^2 = (4^n - 4^(n-1))((4^(n+2) - 4^(n+1))`

Squaring to the left yields:

`4^(2(n+1)) - 2*4^(n+1)*4^n + 4^(2n) = 4^n*4^(n+2) - 4^n*4^(n+1) - 4^(n-1)*4^(n+2) + 4^(n-1)*4^(n+1)`

Using the rules of exponentiation yields:

`4^(2(n+1)) - 2*4^(n+1+n) + 4^(2n) = 4^(n+2+n) - 4^(n+1+n) - 4^(n-1+n+2) + 4^(n-1+n+1)`

`4^(2n+2) - 2*4^(2n+1) + 4^(2n) = 4^(2n+2) - 4^(2n+1) - 4^(2n+1) + 4^(2n)`

`4^(2n+2) - 2*4^(2n+1) + 4^(2n) = 4^(2n+2) - 2*4^(2n+1) + 4^(2n)`

**Since both sides are equal, the validity of relation `(a_n)^2 = a_(n-1)*a_(n+1)` is tested and the given summation `S_n = 4^(n+1) - 1` represents the summation of terms of a geometric sequence.**

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