Let 4^n - 1 represent the sum of m terms of a GP. 4^(n - 1) - 1 is the sum of m - 1 terms of the GP.

Tn = 4^n - 1 - 4^(n - 1) + 1

=> 4^n - 4^n/4

For two consecutive terms Tn and...

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Let 4^n - 1 represent the sum of m terms of a GP. 4^(n - 1) - 1 is the sum of m - 1 terms of the GP.

Tn = 4^n - 1 - 4^(n - 1) + 1

=> 4^n - 4^n/4

For two consecutive terms Tn and Tn+1

Tn+1 / Tn = [4^(n + 1) - 4^(n + 1)/4]/(4^n - 4^n/4)

=> [4*4^n - 4*4^n/4]/(4^n - 4^n/4)

=> 4*(4^n - 4^n/4)/(4^n - 4^n/4)

=> 4

**As we arrive at a common ratio between consecutive terms, 4^n - 1 represents the sum of terms of a GP.**