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Determine the value of the infinite sum \(\sum_{n = 17}^\infty \frac{\binom{n}{15}}{\binom{n}{17}}\)

 Dec 2, 2018
 #1
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∑[(nC15 / nC17), n, 17, infinity]=272

 Dec 2, 2018
 #2
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\(\displaystyle\sum_{n = 17}^\infty \frac{\binom{n}{15}}{\binom{n}{17}}\\ \displaystyle\sum_{n = 17}^\infty \;\;\frac{n!}{15!(n-15)!}\cdot \frac{17!(n-17)!}{n!}\\ \displaystyle\sum_{n = 17}^\infty \;\;\frac{17!(n-17)!}{15!(n-15)!}\\ \displaystyle\sum_{n = 17}^\infty \;\;\frac{16*17}{(n-15)(n-16)}\\ \)

 

\(\displaystyle\sum_{n = 17}^\infty \;\;\frac{16*17}{(n-16)(n-15)}\\ =272\left[\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}+.....\right] \)

 

 

\(Let\\ \frac{1}{(n-16)(n-15)}=\frac{A}{n-16}+\frac{B}{n-15}\\ \frac{1}{(n-16)(n-15)}=\frac{nA-15A}{(n-16)(n-15)}+\frac{nB-16B}{(n-16)(n-15)}\\ \frac{1}{(n-16)(n-15)}=\frac{n(A+B)-(15A+16B)}{(n-16)(n-15)}\\ A+B=0\;\;\;so\;\; B=-A\\ 15A+16B=-1\\ 15A-16A=-1\\ -A=-1\\ A=1 \qquad B=-1\\ so\\ \frac{1}{(n-16)(n-15)}=\frac{1}{n-16}-\frac{1}{n-15}\\\)

 

 

\(\displaystyle\sum_{n = 17}^\infty \;\;\frac{1}{(n-16)(n-15)}\\ \displaystyle\sum_{n = 17}^\infty \;\;\frac{1}{n-16}-\frac{1}{n-15}\\ =\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}....\\ =1\)

 

SO

 

\(\sum_{n = 17}^\infty \frac{\binom{n}{15}}{\binom{n}{17}}=272*1 = 272\)

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 Dec 3, 2018

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