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what is the sum of entries in row 10 of Pascal trigangle

Guest Apr 4, 2017
 #1
avatar+87301 
+1

The sum of the entries in row 10 of Pascal's Triangle  = 2^10  = 1024  where row "0"  = 1

 

 

cool cool cool

CPhill  Apr 4, 2017
 #2
avatar+19639 
+2

what is the sum of entries in row 10 of Pascal trigangle

 

\(\begin{array}{|rcll|} \hline (1+x)^{10} &=& \binom{10}{0} + \binom{10}{1}x \\ &+& \binom{10}{2}x^2 + \binom{10}{3}x^3 \\ &+& \binom{10}{4}x^4 + \binom{10}{5}x^5 \\ &+& \binom{10}{6}x^6 + \binom{10}{7}x^7 \\ &+& \binom{10}{8}x^8 + \binom{10}{9}x^9 \\ &+& \binom{10}{10}x^{10} \\ \hline \end{array} \)

 

Set x = 1:

\(\begin{array}{|rcll|} \hline (1+1)^{10} &=& \binom{10}{0} + \binom{10}{1}\cdot 1 \\ &+& \binom{10}{2}\cdot 1^2 + \binom{10}{3}\cdot 1^3 \\ &+& \binom{10}{4}\cdot 1^4 + \binom{10}{5}\cdot 1^5 \\ &+& \binom{10}{6}\cdot 1^6 + \binom{10}{7}\cdot 1^7 \\ &+& \binom{10}{8}\cdot 1^8 + \binom{10}{9}\cdot 1^9 \\ &+& \binom{10}{10}\cdot 1^{10} \\\\ 2^{10} &=& \binom{10}{0} + \binom{10}{1} + \binom{10}{2} + \binom{10}{3} + \binom{10}{4} + \binom{10}{5} \\ &+& \binom{10}{6} + \binom{10}{7} + \binom{10}{8} + \binom{10}{9} + \binom{10}{10} \\ \hline \end{array} \)

 

The sum of entries in row 10 of Pascal trigangle is 210

 

laugh

heureka  Apr 5, 2017

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