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I am attempting to simply the following calculation where the initial number is $2.4$ and add a decrease of $10$% from the previous step:

$1.$ $2.4\times 0.9 = 2.16$

$2.$ $2.4\times\left(0.9 + {0.9}^2\right) = 4.104$

$3.$ $2.4\times\left(0.9 + {0.9}^2 + {0.9}^3\right) = 5.8536$

$4.$ $2.4\times\left(0.9 + {0.9}^2 + {0.9}^3 + {0.9}^4\right) = 7.42824$

$5.$ $2.4\times\left(0.9 + {0.9}^2 + {0.9}^3 + {0.9}^4 + {0.9}^5\right) = 8.845416$

$\dots$ and so on

How could I simply this pattern when the number of adding is X?

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  • $\begingroup$ It is very unclear what the question is $\endgroup$ – Klangen Sep 11 '19 at 8:02
  • $\begingroup$ trying to create a formula where the increasing number of additional power ofs added on is X $\endgroup$ – Curiosa Sep 11 '19 at 8:10
  • $\begingroup$ This is a simple application of a finite geometric series. $\endgroup$ – Toby Mak Sep 11 '19 at 8:14
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As a hint, consider the geometric series $$a(1+q+q^2+\ldots+q^{n-1}) = a\frac{q^n-1}{q-1}.$$

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