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I could use help simplifying this to find n, thanks!

 Oct 12, 2019
 #1
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+1

simplify {n/2*(2*(3/4) + (n -1)*(3/8)) - 3/4*((1/2)^n) / (1/ - 1)} >=200

 

3 (n^2 + 3 n + 2^(2 - n))>=3200


n<=root of 2^x (3 x^2 + 9 x - 3200) + 12 near x = -8.00368


n>=root of 2^x (3 x^2 + 9 x - 3200) + 12 near x = 31.1943

 Oct 12, 2019
 #2
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CORRECTION OF ABOVE. SORRY ABOUT THAT:

 

simplify {n/2*(2*(3/4) + (n -1)*(3/8)) - [3/4*((1/2)^n) / (1/2 - 1)]} >=200

 

3 (n^2 + 3 n + 2^(3 - n))>=3200


n<=root of 2^x (3 x^2 + 9 x - 3200) + 24 near x = -7.02021


n>=root of 2^x (3 x^2 + 9 x - 3200) + 24 near x = 31.1943

Guest Oct 12, 2019
 #3
avatar+105700 
+1

 

 

 

 

\(\frac{n}{2}((2(\frac{3}{4})+(n-1)(\frac{3}{8}))-\frac{\frac{3}{4}((\frac{1}{2})^n-1)}{\frac{1}{2}-1}\ge200\\ \frac{n}{2}(\frac{3}{2}+(n-1)(\frac{3}{8}))-\frac{\frac{3}{4}((\frac{1}{2})^n-1)}{-\frac{1}{2}}\ge200\\ \frac{n}{2}(\frac{3}{2}+(n-1)(\frac{3}{8}))+2*\frac{3}{4}((\frac{1}{2})^n-1)\ge200\\ \frac{n}{2}(\frac{3}{2}+(n-1)(\frac{3}{8}))+\frac{3}{2}((\frac{1}{2})^n-1)\ge200\\ \frac{16n}{2}(\frac{3}{2}+(n-1)(\frac{3}{8}))+\frac{16*3}{2}((\frac{1}{2})^n-1)\ge3200\\ \frac{8n}{1}(\frac{3}{2}+(n-1)(\frac{3}{8}))+\frac{8*3}{1}((\frac{1}{2})^n-1)\ge3200\\ 12n+3n(n-1)+24*(\frac{1}{2})^n-24\ge3200\\ 12n+3n^2-3n+24*(\frac{1}{2})^n\ge3224\\ 9n+3n^2+24*(\frac{1}{2})^n\ge3224\\ 24*(\frac{1}{2})^n\ge -3n^2-9n+3224\\ \frac{1}{2^n}\ge \frac{ -3n^2-9n+3224}{24}\\ \)

 

https://www.desmos.com/calculator/hlvyuj07cz

 

 

 

 

\(\text{Approximate answer}\\ n\ge 31.316\qquad or \qquad n\le -7.031 \)

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 Oct 13, 2019

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