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Prove that 2^k(n,0)(n,k)-2^(k-1)(n,1)(n-...

Prove that `2^k(n,0)(n,k)-2^(k-1)(n,1)(n-1,k-1)+2^(k-2)(n,2)(n-2,k-2)-.....+(-1)^k(n,k)(n-k,0)=(n,k)`.

Text Solution

Verified by Experts

To show that
`2^(k).""^(n)C_(0).""^(n)C_(k)-2^(k-1).""^(n)C_(1).""^(n-1)C_(k-1)+2^(k-2).""^(n)C_(2).""^(n)C_(k-2)-...+(-1)^(k)""^(n)C_(k)""^(n-k)C_(0)=""^(n)C_(k)`
Taking LHS
`2^(k).""^(n)C_(0).""^(n)C_(k)-2^(k-1).""^(n)C_(1).""^(n-1)C_(k-1) +... +(-1)^(k)""^(n)C_(k).""^(n-k)C_(0)`
`=sum_(r=0)^(k)(-1)^(r).2^(k-r).""^(n)C_(r)""^(n-r)C_(k-r)`
`= sum _(r=0)^(k)(-1)^(r)2^(k-r).(n!)/(r!(n-r)!).((n-r)!)/((k-r)!(n-k)!)`
`= sum _(r=0)^(k)(-1)^(r).2^(k-r).(n!)/(r!(n-r)!k!).(k!)/(r!(k-r)!)`
`= sum _(r=0)^(k)(-1)^(r).2^(k-r). ""^(n)C_(k).""^(n)C_(r)=2^(k).""^(n)C_(k){sum _(r=0)^(k)(-1)^(r).(1)/(2^(r))""^(k)C_(r)}`
`=2^(k).""^(n)C_(k)(1-(1)/(2))^(k)=""^(n)C_(k)=RHS `
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