Binomial Theorem

**Arbitur** · March 27th 2006, 02:33 PM

14 a and b i cant get... please help

**ThePerfectHacker** · March 27th 2006, 02:43 PM

Originally Posted by Arbitur

14 a and b i cant get... please help

$(x+y)^{10}$
by binomial theorem
$ \left(\begin{array}{c}10\\0\end{array}\right)x^{10 }y^0 + \left(\begin{array}{c}10\\1\end{array}\right)x^9y + \left(\begin{array}{c}10\\2\end{array}\right)x^8y^ 2 +$
$ + \left(\begin{array}{c}10\\3\end{array}\right)x^7y^ 3 + \left(\begin{array}{c}10\\4\end{array}\right)x^6y^ 4 + \left(\begin{array}{c}10\\5\end{array}\right)x^5y^ 5 $
$ + \left(\begin{array}{c}10\\6\end{array}\right)x^4y^ 6 + \left(\begin{array}{c}10\\7\end{array}\right)x^3y^ 7 + \left(\begin{array}{c}10\\8\end{array}\right)x^2y^ 8 $ $ + \left(\begin{array}{c}10\\9\end{array}\right)xy^9 + \left(\begin{array}{c}10\\10\end{array}\right)y^{1 0} $
---------
When the combinatorics are evaluated we have,
$x^{10}+10x^9y+45x^8y^2+120x^7y^3+210x^6y^4+$ $252x^5y^5+210x^4y^6+120x^3y^7+45x^8y^2+10x^9y+y^{1 0}$

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