Question

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Answer 1

Let us find some values first:

$\quad ^{n+3}C_{n+1}-^{n+2}C_{n}$

$=\frac{(n+3)!}{(n+1)!2!}-\frac{(n+2)!}{n!2!}$

$=\frac{(n+2)!}{n!2!}\big(\frac{n+3}{n+1}-1\big)$

$=\frac{(n+2)!}{n!2!}.\frac{2}{n+1}$

$=n+2$

Similarly,

$\quad\quad ^{n+4}C_{n+2}-^{n+2}C_{n+1}=n+3$

$\quad\quad ^{n+5}C_{n+3}-^{n+4}C_{n+2}=n+4$

$\quad\quad ^{n+4}C_{n+2}-^{n+2}C_{n}=2n+5$

$\quad\quad ^{n+5}C_{n+3}-^{n+3}C_{n+1}=2n+7$

$\quad\quad ^{n+6}C_{n+4}-^{n+4}C_{n+2}=2n+9$

$\therefore \left|\begin{array}{ccc} ^{n+2}C_{n}&^{n+3}C_{n+1}&^{n+4}C_{n+2}\\ ^{n+3}C_{n+1}&^{n+4}C_{n+2}&^{n+5}C_{n+3}\\ ^{n+4}C_{n+2}&^{n+5}C_{n+3}&^{n+6}C_{n+4}\end{array}\right|$

We do $C'_{2}\to C_{2}-C_{1}\:\&\:C'_{3}\to C_{3}-C_{1}$:

$=\left|\begin{array}{ccc} ^{n+2}C_{n}&^{n+3}C_{n+1}-^{n+2}C_{n}&^{n+4}C_{n+2}-^{n+2}C_{n}\\ ^{n+3}C_{n+1}&^{n+4}C_{n+2}-^{n+3}C_{n+1}&^{n+5}C_{n+3}-^{n+3}C_{n+1}\\ ^{n+4}C_{n+2}&^{n+5}C_{n+3}-^{n+4}C_{n+2}&^{n+6}C_{n+4}-^{n+4}C_{n+2}\end{array}\right|$

$=\left|\begin{array}{ccc} ^{n+2}C_{n}&n+2&2n+5\\ ^{n+3}C_{n+1}&n+3&2n+7\\ ^{n+4}C_{n+2}&n+4&2n+9 \end{array}\right|$

We take $^{n+2}C_{n}$ common from $C_{1}$:

$=^{n+2}C_{n}\left|\begin{array}{ccc}1&n+2&2n+5\\ \frac{n+3}{n+1}&n+3&2n+7\\ \frac{(n+4)(n+3)}{(n+2)(n+1)}&n+4&2n+9\end{array}\right|$

We do $C'_{3}\to C_{3}-2C_{2}$:

$=^{n+2}C_{n}\left|\begin{array}{ccc}1&n+2&1\\ \frac{n+3}{n+1}&n+3&1\\ \frac{(n+4)(n+3)}{(n+2)(n+1)}&n+4&1\end{array}\right|$

We do $R'_{2}\to R_{2}-R_{1}\:\&\:R'_{3}\to R_{3}-R_{1}$:

$=^{n+2}C_{n}\left|\begin{array}{ccc}1&n+2&1\\\frac{2}{n+1}&1&0\\ \frac{4n+10}{(n+2)(n+1)}&2&0\end{array}\right|$

Expanding along $C_{3}$:

$=^{n+2}C_{n}\big[\frac{4}{n+1}-\frac{4n+10}{(n+2)(n+1)}\big]$

$=^{n+2}C_{n}.\frac{4n+8-4n-10}{(n+2)(n+1)}$

$=-^{n+2}C_{n}.\frac{2}{(n+2)(n+1)}$

$=-\frac{(n+2)(n+1)n!}{n!2!}.\frac{2}{(n+2)(n+1)}$

$=-1$

Finally we have :

$\left|\begin{array}{ccc} ^{n+2}C_{n}&^{n+3}C_{n+1}&^{n+4}C_{n+2}\\ ^{n+3}C_{n+1}&^{n+4}C_{n+2}&^{n+5}C_{n+3}\\ ^{n+4}C_{n+2}&^{n+5}C_{n+3}&^{n+6}C_{n+4}\end{array}\right|=-1$

Answer 2

Similarly,

${\quad\quad ^{n+4}C_{n+2}-^{n+2}C_{n+1}=n+3 }$

n+4Cn+2− n+2Cn+1=n+3

${\quad\quad ^{n+5}C_{n+3}-^{n+4}C_{n+2}}$

=n+4n+5Cn+3−n+4Cn+2=n+4

${\quad\quad ^{n+4}C_{n+2}-^{n+2}C_{n}=2n+5 }$

n+4Cn+2−n+2Cn=2n+5

${\quad\quad ^{n+5}C_{n+3}-^{n+3}C_{n+1}}$

=2n+7n+5Cn+3−n+3Cn+1=2n+7

${\quad\quad ^{n+6}C_{n+4}-^{n+4}C_{n+2}}$

=2n+9n+6Cn+4−n+4Cn+2=2n+9

${\begin{gathered}\therefore \left|\begin{array}{ccc} ^{n+2}C_{n}&^{n+3}C_{n+1}&^{n+4}C_{n+2}\\ ^{n+3}C_{n+1}&^{n+4}C_{n+2}&^{n+5}C_{n+3}\\ ^{n+4}C_{n+2}&^{n+5}C_{n+3}&^{n+6}C_{n+4}\end{array}\right|\end{gathered}}$

∴n+2Cnn+3Cn+1n+4Cn+2n+3Cn+1n+4Cn+2n+5Cn+3n+4Cn+2n+5Cn+3n+6Cn+4

${ We do C'_{2}\to C_{2}-C_{1}\:\&\:C'_{3}\to C_{3}-C_{1}C }$

2′→C2−C1&C3′→C3−C1:

${\begin{gathered}=\left|\begin{array}{ccc} ^{n+2}C_{n}&^{n+3}C_{n+1}-^{n+2}C_{n}&^{n+4}C_{n+2}-^{n+2}C_{n}\\ ^{n+3}C_{n+1}&^{n+4}C_{n+2}-^{n+3}C_{n+1}&^{n+5}C_{n+3}-^{n+3}C_{n+1}\\ ^{n+4}C_{n+2}&^{n+5}C_{n+3}-^{n+4}C_{n+2}&^{n+6}C_{n+4}-^{n+4}C_{n+2}\end{array}\right|\end{gathered} }$

${ We \: take ^{n+2}C_{n} }$

n+2Cn

${ common \: from C_{1}C }$

${\begin{gathered}=^{n+2}C_{n}\left|\begin{array}{ccc}1&n+2&2n+5\\ \frac{n+3}{n+1}&n+3&2n+7\\ \frac{(n+4)(n+3)}{(n+2)(n+1)}&n+4&2n+9\end{array}\right|\end{gathered} }$

=n+2Cn1n+1n+3(n+2)(n+1)(n+4)(n+3)n+2n+3n+42n+52n+72n+9

${ We \: do C'_{3}\to C_{3}-2C_{2}C }$

3′→C3−2C2:

${ \begin{gathered}=^{n+2}C_{n}\left|\begin{array}{ccc}1&n+2&1\\ \frac{n+3}{n+1}&n+3&1\\ \frac{(n+4)(n+3)}{(n+2)(n+1)}&n+4&1\end{array}\right|\end{gathered} }$

=n+2Cn

1n+1n+3(n+2)(n+1)(n+4)(n+3)n+2n+3n+4111

${ We \: do \: R'_{2}\to R_{2}-R_{1}\:\&\:R'_{3}\to R_{3}-R_{1}R }$

2′→R2−R1&R3′→R3−R1:

${\begin{gathered}=^{n+2}C_{n}\left|\begin{array}{ccc}1&n+2&1\\\frac{2}{n+1}&1&0\\ \frac{4n+10}{(n+2)(n+1)}&2&0\end{array}\right|\end{gathered}}[tex]</p><p>=n+2Cn1n+12(n+2)(n+1)4n+10n+212100</p><p>[tex]{Expanding \: along C_{3}C3:}$

${ =^{n+2}C_{n}\big[\frac{4}{n+1}-\frac{4n+10}{(n+2)(n+1)}\big]= }$

n+2Cn[ n+14−(n+2)(n+1)4n+10]

${=^{n+2}C_{n}.\frac{4n+8-4n-10}{(n+2)(n+1)}= }$

n+2Cn.(n+2)(n+1)4n+8−4n−10

${ =-^{n+2}C_{n}.\frac{2}{(n+2)(n+1)}=− }$

n+2Cn.(n+2)(n+1)2

${ =-\frac{(n+2)(n+1)n!}{n!2!}.\frac{2}{(n+2)(n+1)}=− </p><p>n!2!}$

(n+2)(n+1)n!.(n+2)(n+1)2=-1=−1

Finally we have :

${\begin{gathered}\left|\begin{array}{ccc} ^{n+2}C_{n}&^{n+3}C_{n+1}&^{n+4}C_{n+2}\\ ^{n+3}C_{n+1}&^{n+4}C_{n+2}&^{n+5}C_{n+3}\\ ^{n+4}C_{n+2}&^{n+5}C_{n+3}&^{n+6}C_{n+4}\end{array}\right|=-1\end{gathered} }$

n+2Cnn+3Cn+1n+4Cn+2

n+3C n+1n+4Cn+2n+5Cn+3n+4Cn+2n+5Cn+3n+6Cn+4=−1n+6Cn+4−n+4Cn+2=2n+9

${\begin{gathered}\therefore \left|\begin{array}{ccc} ^{n+2}C_{n}&^{n+3}C_{n+1}&^{n+4}C_{n+2}\\ ^{n+3}C_{n+1}&^{n+4}C_{n+2}&^{n+5}C_{n+3}\\ ^{n+4}C_{n+2}&^{n+5}C_{n+3}&^{n+6}C_{n+4}\end{array}\right|\end{gathered} }$

∴n+2Cnn+3Cn+1n+4Cn+2n+3Cn+1n+4Cn+2n+5Cn+3n+4Cn+2n+5Cn+3n+6Cn+4

${We \: do C'_{2}\to C_{2}-C_{1}\:\&\:C'_{3}\to C_{3}-C_{1}C }$

2′→C2−C1&C3′→C3−C1 :

${\begin{gathered}=\left|\begin{array}{ccc} ^{n+2}C_{n}&^{n+3}C_{n+1}-^{n+2}C_{n}&^{n+4}C_{n+2}-^{n+2}C_{n}\\ ^{n+3}C_{n+1}&^{n+4}C_{n+2}-^{n+3}C_{n+1}&^{n+5}C_{n+3}-^{n+3}C_{n+1}\\ ^{n+4}C_{n+2}&^{n+5}C_{n+3}-^{n+4}C_{n+2}&^{n+6}C_{n+4}-^{n+4}C_{n+2}\end{array}\right|\end{gathered} }$

=n+2Cnn+3Cn+1n+4Cn+2n+3Cn+1−n+2Cnn+4Cn+2−n+3Cn+1n+5Cn+3−n+4C n+2n+4Cn+2−n+2Cnn+5Cn+3−n+3Cn+1n+6Cn+4−n+Cn+2

${\begin{gathered}=\left|\begin{array}{ccc} ^{n+2}C_{n}&n+2&2n+5\\ ^{n+3}C_{n+1}&n+3&2n+7\\ ^{n+4}C_{n+2}&n+4&2n+9 \end{array}\right|\end{gathered} }$

=n+2Cnn+3Cn+1n+4Cn+2n+2n+3n+42n+52n+72n+9

${We\: take ^{n+2}C_{n} }$n+2Cn

${common from C_{1}C1 : }$

${\begin{gathered}=^{n+2}C_{n}\left|\begin{array}{ccc}1&n+2&2n+5\\ \frac{n+3}{n+1}&n+3&2n+7\\ \frac{(n+4)(n+3)}{(n+2)(n+1)}&n+4&2n+9\end{array}\right|\end{gathered} }$

=n+2Cn1n+n+3(n+2)(n+1)(n+4)(n+3)n+2n+3n+42n+52n+72n+9

${We \: do C'_{3}\to C_{3}-2C_{2}C }$

3′→C3−2C2 :

${\begin{gathered}=^{n+2}C_{n}\left|\begin{array}{ccc}1&n+2&1\\ \frac{n+3}{n+1}&n+3&1\\ \frac{(n+4)(n+3)}{(n+2)(n+1)}&n+4&1\end{array}\right|\end{gathered} }$

=n+2Cn1n+1n+3(n+2)(n+1)(n+4)(n+3)n+2n+3n+4111

${We \: do \: R'_{2}\to R_{2}-R_{1}\:\&\:R'_{3}\to R_{3}-R_{1}R }$

2′→R2−R1&R3′→R3−R1 :

${\begin{gathered}=^{n+2}C_{n}\left|\begin{array}{ccc}1&n+2&1\\\frac{2}{n+1}&1&0\\ \frac{4n+10}{(n+2)(n+1)}&2&0\end{array}\right|\end{gathered} }$

=n+2Cn1n+12(n+2)(n+1)4n+10n+212100

${ Expanding \: along C_{3}C3:}$

${=^{n+2}C_{n}\big[\frac{4}{n+1}-\frac{4n+10}{(n+2)(n+1)}\big]= }$n+2Cn[n+14−(n+2)(n+1)4n+10]

${=^{n+2}C_{n}.\frac{4n+8-4n-10}{(n+2)(n+1)}= }$n+2Cn.(n+2)(n+1)4n+8−4n−10

${=-^{n+2}C_{n}.\frac{2}{(n+2)(n+1)}=−}$

n+2Cn.(n+2)(n+1)2

${=-\frac{(n+2)(n+1)n!}{n!2!}.\frac{2}{(n+2)(n+1)}=− </p><p>n!2!}$(n+2)(n+1)n!.(n+2)(n+1)2=-1=−1