# What is the smallest integer from sum: 1.04 + 1.04^3 + 1.04^5 + 1.04^7 +....+1.04^151can use calculator

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to determine if the terms of the sum are the terms of a geometric series using the following property such that:

`q = (a_(n+1))/(a_n)`

Substituting `1.04^3`  for `a_(n+1)`  and 1.04 for `a_n`  yields:

`q = (1.04^3)/(1.04) =gt q = 1.04^2`

You should verify if `q=1.04^2`  is the common ratio, hence, you should multiply `1.04^3`  by `1.04^2`  to check if the result is the next term of the sum:

`1.04^3*1.04^2 = 1.04^(3+2) = 1.04^5`

Notice, that the result represents th next term, hence, you need to find the sum of a number of terms of a geometric series using the following formula such that:

`s_n = 1.04*(1.04^n - 1)/(1.04-1)`

Notice that you don't know how many terms does the sum have, hence, you should find n using the formula of general term such that:

`a_n = a_1*q^(n-1)`

Substituting `1.04^151 `  for `a_n, 1.04^2`  for q and 1.04 for `a_1`  yields:

`1.04^151 = 1.04*1.04^(2(n-1))`

`1.04^151 = 1.04^(2n - 2 + 1)`

`1.04^151 = 1.04^(2n - 1) =gt 151 = 2n - 1 =gt 2n = 152 =gt n = 76`

Hence, there are 76 terms and you may evaluate the sum, substituting 76 for n in formula such that:

`s_n = 1.04*(1.04^76 - 1)/(1.04-1)`

Hence, evaluating the given sum yields `s_n = 1.04*(1.04^76 - 1)/(1.04-1).`

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