If n is a positive integer, prove that r=1∑nr3( nCr−1 nCr ) 2=12n(n+1) 2(n+2) .

nCr=(n−r)!.r!n!=(n−(r−1))(n−r)!.(r−1)!.r(n−(r−1))n!=(n−(r−1)!)(r−1)!.r(n−(r−1))n!=rn−(r−1)nCr−1Hence, nCr−1nCr=rn+1−rHence, (nCr−1nCr)2=r2(n+1−r)2r3(nCr−1nCr)2=r(n+1−r)2=r[(n+1)2−2r(n+1)+r2]=r(n+1)2−2r2(n+1)+r3∑r3(nCr−1nCr)2=∑r(n+1)2−2r2(n+1)+r3=(n+1)2∑r−2(n+1)∑r2+∑r3=(n+1)2(2n(n+1))−(2(n+1))(6n(n+1)(2n+1))+4n2(n+1)2=2n(n+1)3−62n(n+1)2(2n+1)+4n2(n+1)2=2n(n+1)2[(n+1)−32(2n+1)+2n]=2n(n+1)2[66n+6−8n−4+3n]=2n(n+1)2[6n+2]=12n(n+1)2(n+2)

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If $n$ is a positive integer, prove that $r = 1 \sum n r^{3} (\frac{_{n} C _{r}}{_{n} C _{r - 1}})^{2} = \frac{n ( n + 1 ) ^{2} ( n + 2 )}{12}$ .

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n

is a positive integer, prove that

r = 1 \sum n r^{3} (\frac{_{n} C _{r}}{_{n} C _{r - 1}})^{2} = \frac{n ( n + 1 ) ^{2} ( n + 2 )}{12}

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^{n} C_{r} = \frac{n !}{( n - r ) ! . r !}

= \frac{( n - ( r - 1 ) ) n !}{( n - ( r - 1 ) ) ( n - r ) ! . ( r - 1 ) ! . r}

= \frac{( n - ( r - 1 ) ) n !}{( n - ( r - 1 ) ! ) ( r - 1 ) ! . r}

= \frac{n - ( r - 1 )}{r}^{n} C_{r - 1}

Hence,

\frac{^{n} C _{r}}{^{n} C _{r - 1}} = \frac{n + 1 - r}{r}

Hence,

(\frac{^{n} C _{r}}{^{n} C _{r - 1}})^{2} = \frac{( n + 1 - r ) ^{2}}{r ^{2}}

r^{3} (\frac{^{n} C _{r}}{^{n} C _{r - 1}})^{2} = r (n + 1 - r)^{2}

= r [(n + 1)^{2} - 2 r (n + 1) + r^{2}]

= r (n + 1)^{2} - 2 r^{2} (n + 1) + r^{3}

\sum r^{3} (\frac{^{n} C _{r}}{^{n} C _{r - 1}})^{2} = \sum r (n + 1)^{2} - 2 r^{2} (n + 1) + r^{3}

= (n + 1)^{2} \sum r - 2 (n + 1) \sum r^{2} + \sum r^{3}

= (n + 1)^{2} (\frac{n ( n + 1 )}{2}) - (2 (n + 1)) (\frac{n ( n + 1 ) ( 2 n + 1 )}{6}) + \frac{n ^{2} ( n + 1 ) ^{2}}{4}

= \frac{n ( n + 1 ) ^{3}}{2} - \frac{2 n ( n + 1 ) ^{2} ( 2 n + 1 )}{6} + \frac{n ^{2} ( n + 1 ) ^{2}}{4}

= \frac{n ( n + 1 ) ^{2}}{2} [(n + 1) - \frac{2 ( 2 n + 1 )}{3} + \frac{n}{2}]

= \frac{n ( n + 1 ) ^{2}}{2} [\frac{6 n + 6 - 8 n - 4 + 3 n}{6}]

= \frac{n ( n + 1 ) ^{2}}{2} [\frac{n + 2}{6}]

= \frac{n ( n + 1 ) ^{2} ( n + 2 )}{12}

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Question Text	If $n$ is a positive integer, prove that $r = 1 \sum n r^{3} (\frac{_{n} C _{r}}{_{n} C _{r - 1}})^{2} = \frac{n ( n + 1 ) ^{2} ( n + 2 )}{12}$ .
Answer Type	Text solution:1
Upvotes	150

If n is a positive integer, prove that r=1∑n​r3( n​Cr−1​ n​Cr​​ ) 2=12n(n+1) 2(n+2) ​.

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If $n$ is a positive integer, prove that $r = 1 \sum n r^{3} (\frac{_{n} C _{r}}{_{n} C _{r - 1}})^{2} = \frac{n ( n + 1 ) ^{2} ( n + 2 )}{12}$ .