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Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 8, 2011"

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By the Binomial Theorem, we have  
 
By the Binomial Theorem, we have  
  
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<math> \begin {align*} \left(a-\frac{1}{\sqrt{a}}\right)^{7} &= a^7 + \dbinom {7}{1} (a^6) \left( - \frac {1}{\sqrt {a}} \right) + \dbinom {7}{2} (a^5) \left( - \frac {1}{\sqrt {a}} \right)^2 + \dbinom {7}{3} (a^4) \left( - frac {1}{\sqrt {a}} \right)^3 + \dbinom {7}{4} (a^3) \left( - \frac {1}{\sqrt {a}} \right)^4 + \dbinom {7}{5} (a^2) \left( - \frac {1}{\sqrt {a}} \right)^5 + 7 (a) \left( - \frac {1}{\sqrt {a}} \right)^6 + \left( - \frac {1}{\sqrt {a}} \right)^7 \\&= a^7 - 7a^{11/2} + 21 a^4 - 35 a^{5/2} + 35 a - 21 a^{-1/2} + 7a^{-3} - a^{-7/2} </math> So, the answer is <math> \boxed {-21} </math>.
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<math> \left(a-\frac{1}{\sqrt{a}}\right)^{7} = a^7 + \dbinom {7}{1} (a^6) \left( - \frac {1}{\sqrt {a}} \right) + \dbinom {7}{2} (a^5) \left( - \frac {1}{\sqrt {a}} \right)^2 + \dbinom {7}{3} (a^4) \left( - \frac {1}{\sqrt {a}} \right)^3 + \dbinom {7}{4} (a^3) \left( - \frac {1}{\sqrt {a}} \right)^4 + \dbinom {7}{5} (a^2) \left( - \frac {1}{\sqrt {a}} \right)^5 + 7 (a) \left( - \frac {1}{\sqrt {a}} \right)^6 + \left( - \frac {1}{\sqrt {a}} \right)^7 </math>
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<math> = a^7 - 7a^{11/2} + 21 a^4 - 35 a^{5/2} + 35 a - 21 a^{-1/2} + 7a^{-3} - a^{-7/2} </math> So, the answer is <math> \boxed {-21} </math>.

Revision as of 01:33, 8 July 2011

Problem

AoPSWiki:Problem of the Day/July 8, 2011

Solution

This Problem of the Day needs a solution. If you have a solution for it, please help us out by adding it.

By the Binomial Theorem, we have

$\left(a-\frac{1}{\sqrt{a}}\right)^{7} = a^7 + \dbinom {7}{1} (a^6) \left( - \frac {1}{\sqrt {a}} \right) + \dbinom {7}{2} (a^5) \left( - \frac {1}{\sqrt {a}} \right)^2 + \dbinom {7}{3} (a^4) \left( - \frac {1}{\sqrt {a}} \right)^3 + \dbinom {7}{4} (a^3) \left( - \frac {1}{\sqrt {a}} \right)^4 + \dbinom {7}{5} (a^2) \left( - \frac {1}{\sqrt {a}} \right)^5 + 7 (a) \left( - \frac {1}{\sqrt {a}} \right)^6 + \left( - \frac {1}{\sqrt {a}} \right)^7$

$= a^7 - 7a^{11/2} + 21 a^4 - 35 a^{5/2} + 35 a - 21 a^{-1/2} + 7a^{-3} - a^{-7/2}$ So, the answer is $\boxed {-21}$.

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