Laws of summations

Algebra Level 4

For all integer k > 1 k > 1

∑ r = 1 k 2 r ( n r ) ( k − 1 r − 1 ) ∑ r = 0 k ( n r ) ( n + k − r − 1 n − 1 ) = ? \large \frac { \displaystyle \sum _{ r=1 }^{ k }{ { 2 }^{ r }\begin{pmatrix} n \\ r \end{pmatrix}\begin{pmatrix} k-1 \\ r-1 \end{pmatrix} } }{\displaystyle \sum _{ r=0 }^{ k }{ \begin{pmatrix} n \\ r \end{pmatrix}\begin{pmatrix} n+k-r-1 \\ n-1 \end{pmatrix} } }=?


The answer is 1.

Priyanshu Mishra by Priyanshu Mishra , 20, India

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1 solution

Let us analyse the numerator :

N = ∑ r = 1 k 2 r ( n r ) ( k − 1 k − r ) N= \displaystyle\sum_{r=1}^k 2^r \dbinom{n}{r}\dbinom{k-1}{k-r}

Above expansion is equal to the coefficient of x k x^k in the expansion of :

( 1 + 2 x ) n ( 1 + x ) k − 1 = ( 1 + x + x ) n ( 1 + x ) k − 1 (1+2x)^n(1+x)^{k-1}= (1+x+x)^n(1+x)^{k-1}

= ∑ r = 0 n ( n r ) x r ( 1 + x ) n − r ( 1 + x ) k − 1 = \displaystyle\sum_{r=0}^n \dbinom{n}{r} x^r (1+x)^{n-r}(1+x)^{k-1}

= ∑ r = 0 n ( n r ) x r ( 1 + x ) n + k − r − 1 = \displaystyle\sum_{r=0}^n \dbinom{n}{r} x^r (1+x)^{n+k-r-1}

Coefficient of x k x^k in the above expression is :

∑ r = 0 n ( n r ) ( n + k − r − 1 k − r ) = ∑ r = 0 n ( n r ) ( n + k − r − 1 n − 1 ) \displaystyle\sum_{r=0}^n \dbinom{n}{r}\dbinom{n+k-r-1}{k-r}=\displaystyle\sum_{r=0}^n \dbinom{n}{r}\dbinom{n+k-r-1}{n-1}

Hence ,

∑ r = 0 n ( n r ) ( n + k − r − 1 n − 1 ) = ∑ r = 1 k 2 r ( n r ) ( k − 1 r − 1 ) \displaystyle\sum_{r=0}^n \dbinom{n}{r}\dbinom{n+k-r-1}{n-1} = \displaystyle\sum_{r=1}^k 2^r \dbinom{n}{r}\dbinom{k-1}{r-1}

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You're missing 1 small final step. Do you see what it is?

Calvin Lin Staff - 3 years, 3 months ago

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