Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 8, 2011"
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| + | By the Binomial Theorem, we have <cmath> \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} </cmath> So, the answer is <math> \boxed {-21} </math>. | ||
Revision as of 01:26, 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
\[\begin {align*} \left(a-\frac{1}{\sqrt{a}}\right)^{7} &= a^7 + \dbinom {7}{1} (a^6) \left( - \frac {1}{\sqrt {a}} \right) + \dbinom {7}{2} (a^5) \left( - \frac {1}{\sqrt {a}} \right)^2 + \dbinom {7}{3} (a^4) \left( - frac {1}{\sqrt {a}} \right)^3 + \dbinom {7}{4} (a^3) \left( - \frac {1}{\sqrt {a}} \right)^4 + \dbinom {7}{5} (a^2) \left( - \frac {1}{\sqrt {a}} \right)^5 + 7 (a) \left( - \frac {1}{\sqrt {a}} \right)^6 + \left( - \frac {1}{\sqrt {a}} \right)^7 \\&= a^7 - 7a^{11/2} + 21 a^4 - 35 a^{5/2} + 35 a - 21 a^{-1/2} + 7a^{-3} - a^{-7/2}\] (Error compiling LaTeX. Unknown error_msg) So, the answer is 
.
