Binomial coefficients

Probability Level 1

2 903 ∑ n = 0 451 ( 903 n ) = ? \frac{2^{903}}{ \displaystyle \sum_{n=0}^{451} \binom{903}n } = \, ?


Notation : ( M N ) \binom MN denotes the binomial coefficient , ( M N ) = M ! N ! ( M − N ) ! \binom MN = \frac{M!}{N!(M-N)!} .


The answer is 2.

Geoff Pilling by Geoff Pilling , 53, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Geoff Pilling
May 6, 2016

2 903 = ∑ n = 0 903 ( 903 n ) = ∑ n = 0 451 ( 903 n ) + ∑ n = 452 903 ( 903 n ) = 2 ∑ n = 0 451 ( 903 n ) 2^{903} = \sum_{n=0}^{903}\binom{903}{n}= \sum_{n=0}^{451}\binom{903}{n} + \sum_{n=452}^{903}\binom{903}{n} = 2\sum_{n=0}^{451}\binom{903}{n} (by symmetry)

Therefore,

2 903 ∑ n = 0 451 ( 903 n ) = 2 \mathbf{\frac{2^{903}}{\sum_{n=0}^{451}\binom{903}{n}}}=\boxed2

,

14 Helpful 2 Interesting 3 Brilliant 2 Confused

Geoff, how would one realize the first equality?

Malthe Christensen - 3 years, 10 months ago

Log in to reply

This is known as the Sum of Binomial Coefficients over Lower Index

Geoff Pilling - 3 years, 10 months ago

http://mathforum.org/mathimages/index.php/Pascal's triangle#Sum of_Rows

Rachel Laubacher - 3 years, 4 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...