Binohow?

Algebra Level 4

∑ r = 0 n r + 1 ( n r ) = ∑ r = 0 n n 2 − 3 n + 5 2 ( n r ) \large \displaystyle \sum_{r=0}^n \frac{r + 1}{{n \choose r}} = \displaystyle \sum_{r=0}^n \frac{n^2 -3n +5 }{2{n \choose r}}

If n 1 n_{1} and n 2 n_{2} are the two values of n n which satisfy the given equation above, find the value of n 1 2 + n 2 2 n_{1}^2 + n_{2}^2 .


The answer is 10.

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Ankush Tiwari by Ankush Tiwari , 21, India

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1 solution

Abhishek Sharma
Apr 21, 2015

Consider ∑ r = 0 n r n C r \large \sum _{ r=0 }^{ n }{ \frac { r }{ { ^{ n }{ C } }_{ r } } } .

Using the result ∑ r = 0 n f ( r ) = ∑ r = 0 n f ( n − r ) \sum _{ r=0 }^{ n }{ f(r) } =\sum _{ r=0 }^{ n }{ f(n-r) } we get, ∑ r = 0 n r n C r = ∑ r = 0 n n − r n C n − r \sum _{ r=0 }^{ n }{ \frac { r }{ { ^{ n }{ C } }_{ r } } } =\sum _{ r=0 }^{ n }{ \frac { n-r }{ { ^{ n }{ C } }_{ n-r } } } Also we have n C n − r = n C r \large ^{ n }{ C } _{ n-r }=^{ n }{ C } _{ r } . ∑ r = 0 n r n C r = ∑ r = 0 n n − r n C r \sum _{ r=0 }^{ n }{ \frac { r }{ { ^{ n }{ C } }_{ r } } } =\sum _{ r=0 }^{ n }{ \frac { n-r }{ { ^{ n }{ C } }_{ r } } }

∑ r = 0 n r n C r = ∑ r = 0 n n n C r − ∑ r = 0 n r n C r \sum _{ r=0 }^{ n }{ \frac { r }{ { ^{ n }{ C } }_{ r } } } =\sum _{ r=0 }^{ n }{ \frac { n }{ { ^{ n }{ C } }_{ r } } } -\sum _{ r=0 }^{ n }{ \frac { r }{ { ^{ n }{ C } }_{ r } } } ∑ r = 0 n r n C r = 1 2 ∑ r = 0 n n n C r \sum _{ r=0 }^{ n }{ \frac { r }{ { ^{ n }{ C } }_{ r } } } =\frac { 1 }{ 2 } \sum _{ r=0 }^{ n }{ \frac { n }{ { ^{ n }{ C } }_{ r } } } ∑ r = 0 n r n C r = n 2 ∑ r = 0 n 1 n C r \sum _{ r=0 }^{ n }{ \frac { r }{ { ^{ n }{ C } }_{ r } } } =\frac { n }{ 2 } \sum _{ r=0 }^{ n }{ \frac { 1 }{ { ^{ n }{ C } }_{ r } } } ∑ r = 0 n r n C r + ∑ r = 0 n 1 n C r = n 2 ∑ r = 0 n 1 n C r + ∑ r = 0 n 1 n C r \sum _{ r=0 }^{ n }{ \frac { r }{ { ^{ n }{ C } }_{ r } } } +\sum _{ r=0 }^{ n }{ \frac { 1 }{ { ^{ n }{ C } }_{ r } } } =\frac { n }{ 2 } \sum _{ r=0 }^{ n }{ \frac { 1 }{ { ^{ n }{ C } }_{ r } } } +\sum _{ r=0 }^{ n }{ \frac { 1 }{ { ^{ n }{ C } }_{ r } } } ∑ r = 0 n r + 1 n C r = n + 2 2 ∑ r = 0 n 1 n C r \sum _{ r=0 }^{ n }{ \frac { r+1 }{ { ^{ n }{ C } }_{ r } } } =\frac { n+2 }{ 2 } \sum _{ r=0 }^{ n }{ \frac { 1 }{ { ^{ n }{ C } }_{ r } } } ∑ r = 0 n r + 1 n C r = n + 2 2 ∑ r = 0 n 1 n C r = n 2 − 3 n + 5 2 ∑ r = 0 n 1 n C r \sum _{ r=0 }^{ n }{ \frac { r+1 }{ { ^{ n }{ C } }_{ r } } } =\frac { n+2 }{ 2 } \sum _{ r=0 }^{ n }{ \frac { 1 }{ { ^{ n }{ C } }_{ r } } } =\frac { { n }^{ 2 }-3n+5 }{ 2 } \sum _{ r=0 }^{ n }{ \frac { 1 }{ { ^{ n }{ C } }_{ r } } } n + 2 2 = n 2 − 3 n + 5 2 \frac { n+2 }{ 2 } =\frac { { n }^{ 2 }-3n+5 }{ 2 } n = 1 n=1 n = 3 n=3 10 \boxed{10}

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