How to find n? `50000xx0.05xxn + 50000 lt 50000xx(1.035)^n`

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thilina-g | College Teacher | (Level 1) Educator

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`50000xx0.05xxn + 50000 lt 50000xx(1.035)^n`

Divide both sides by 50000
`1+0.05nlt(1.035)^n`

I can write this as,

`(1+0.035)^ngt(1+0.05n)`

I can use the binomial expansion to expand the left side.

`1+nxx0.035+^nC_2xx0.035^2+^nC_3xx0.035^3gt1+0.05n`

`0.035n+(n(n-1))/2xx0.035^2+(n(n-1)(n-2))/6 xx 0.035^3gt0.05n`

Multiply by 6 and divide by `0.035^3` .

`4897.96n+85.71n(n-1)+n(n-1)(n-2)gt6997.08n`

`n^3-3n^2+2n+85.71n^2-85.71n+4897.96n-6997.08ngt0`

`n^3+82.71n^2-2184.83ngt0`

`n(n^2+82.71n-2184.83)gt0`

`n(n-21.0554)(n+103.7655)gt0` ` `

`n<-103.7655`

`n(n-21.0554)(n+103.7655)<0`

So the inequality doesn't exist.` `

`-103.7655ltnlt0`

`n(n-21.0554)(n+103.7655)gt0`

The inequality is exists.

`0 ltnlt 21.0554`

`n(n-21.0554)(n+103.7655)<0`

So the inequality doesn't exist.

`ngt21.0554`

`n(n-21.0554)(n+103.7655)gt0`

The inequality is exists.

 

Therefore we have two ranges of solutions.

`ngt21.0554`

and

`-103.7655ltnlt0`

 

Actually this is an approximate answer, you can increase the accuracy by increasing the terms of binomial expansion.

 

Sources:

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