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Find the value of the sum $\binom{99}{0} + \binom{99}{2} + \binom{99}{4} + \dots + \binom{99}{98}.$

 May 19, 2021

Best Answer 

 #1
avatar+177 
+2

We can see that if we put in all of the odd terms, we get \(2^{99}\) , however, the odd terms are missing. 

 

But, we see that for each even term \(\binom{99}{2x}\) there is a odd term \(\binom{99}{99-2x}\) which is equal to it.

 

So, the sum is \(2^{98}\).

 May 19, 2021
 #1
avatar+177 
+2
Best Answer

We can see that if we put in all of the odd terms, we get \(2^{99}\) , however, the odd terms are missing. 

 

But, we see that for each even term \(\binom{99}{2x}\) there is a odd term \(\binom{99}{99-2x}\) which is equal to it.

 

So, the sum is \(2^{98}\).

ilovepizza547 May 19, 2021
 #2
avatar+120075 
0

Very nice, Ilovepizza  !!!!

 

 

cool cool cool

CPhill  May 19, 2021
 #3
avatar+25959 
+2

Find the value of the sum
\(\dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4} + \dots + \dbinom{99}{96}+ \dbinom{99}{98}\).

 

\(\text{Formula: } \qquad \dbinom{n}{k}= \dbinom{n}{n-k} \)

 

\(\begin{array}{|rcll|} \hline && \mathbf{ \dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4} + \dots + \dbinom{99}{96}+ \dbinom{99}{98}} \\\\ &=& \dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4} + \dots + \dbinom{99}{48} \\ && +\dbinom{99}{98}+ \dbinom{99}{96}+ \dbinom{99}{94} + \dots +\dbinom{99}{50}\\\\ &=& \dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4}+ \dbinom{99}{6} + \dots + \dbinom{99}{48} \\ && + \dbinom{99}{1}+ \dbinom{99}{3} + \dbinom{99}{5} + \dbinom{99}{7}+ \dots +\dbinom{99}{49}\\\\ &=& \dbinom{99}{0} + \dbinom{99}{1} + \dbinom{99}{2}+ \dbinom{99}{3} + \dbinom{99}{4}+ \dots \\ && + \dbinom{99}{46} + \dbinom{99}{47} + \dbinom{99}{48}+\dbinom{99}{49}\\\\ &=& \dfrac{2^{99}}{2} \\\\ &=& \mathbf{ 2^{98} } \\ \hline \end{array}\)

 

laugh

 May 19, 2021
edited by heureka  May 19, 2021
 #4
avatar+120075 
+1

Thanks, heureka  !!!!

 

 

cool cool cool

CPhill  May 19, 2021
 #5
avatar+113698 
+1

Thanks Ilovepizza and Heureka,

 

Where does 2^99 come from?  I understand everything else. 

 May 19, 2021
 #6
avatar+25959 
+2

Where does \(2^{99}\) come from?

 

Pascal's triangle:


See the sum of the each line:
\(\small{ \begin{array}{|l|rclcl|} \hline n \\ \hline 0 &\dbinom{0}{0} &=& 1 &=& 2^0 \\ 1 &\dbinom{1}{0}+\dbinom{1}{1} &=& 1+1 &=& 2^1 \\ 2 &\dbinom{2}{0}+\dbinom{2}{1}+\dbinom{2}{2} &=& 1+2+1 &=& 2^2 \\ 3 &\dbinom{3}{0}+\dbinom{3}{1}+\dbinom{3}{2}+\dbinom{3}{3} &=& 1+3+3+1 &=& 2^3 \\ 4 &\dbinom{4}{0}+\dbinom{4}{1}+\dbinom{4}{2}+\dbinom{4}{3}+\dbinom{4}{4} &=& 1+4+6+4+1 &=& 2^4 \\ 5 &\dbinom{5}{0}+\dbinom{5}{1}+\dbinom{5}{2}+\dbinom{5}{3}+\dbinom{5}{4}+\dbinom{5}{5} &=& 1+5+10+10+5+1 &=& 2^5 \\ \ldots \\ \color{red}99 & \dbinom{99}{0} + \dbinom{99}{1} + \dots + \dbinom{99}{98}+\dbinom{99}{99} &=&1+99+\dots + 99 + 1 &=& \color{red}2^{99} \\ \hline \end{array} }\)

 

\(\begin{array}{|rcll|} \hline && \color{red}(1+1)^n \\\\ &=&\sum \limits_{k=0}^{n}\dbinom{n}{k} \\ \\ &=& \dbinom{n}{0}+\dbinom{n}{1} + \dots + \dbinom{n}{n-1}+\dbinom{n}{n} \\\\ &=& \color{red}2^n \\ \hline \end{array}\)

 

laugh

 May 20, 2021
 #7
avatar+113698 
0

Thanks very much Heureka.   laugh cool laugh

 

Hopefully I will remember next time.  ://

Melody  May 20, 2021

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