1.

Evaluate sum_(i=0)^(n-1) sum_(j= 1 + i)^(n+1)""^(n)C_(i) ""^(n+1)C_(j) .

Answer»

Solution :Let ` P = sum_(i=0)^(n-1) sum_(j= 1 + i)^(n+1)""^(n)C_(i) ""^(n+1)C_(j)` .
`sum_(j=1)^(n-1)""^(n)C_(0)""^(n+1)C_(j) +sum_(j= 2)^(n+1)""^(n)C_(i) ""^(n+1)C_(j)+ sum_(j= 3)^(n+1) ""^(n)C_(2)""^(n+1)C_(j) + ...+ sum_(j = n)^(n+1) ""^(n)C_(n-1) ""^(n+1)C_(j)`
`""^(n)C_(0) sum_(j=1)^(n-1)""^(n)C_(j)+""^(n)C_(1) +sum_(j= 2)^(n+1)""^(n+1)C_(j) ""^(n+1)C_(2)+ sum_(j= 3)^(n+1) ""^(n+1)C_(j) + ...+ ""^(n)C_(n-1)sum_(j = n)^(n+1)""^(n+1)C_(j)`
`""^(n)C_(0)(""^(n +1)C_(j) + ""^(n+1)C_(2) + ""^(n+1)C_(3) + ...+ ""^(n+1)C_(n+1))`
` + ""^(n)C_(1) (""^(n+1)C_(2)+""^(n+1)C_(3) + ""^(n+1)C_(4)+ ...+ ""^(n+1)C_(n+1))`
` + ""^(n)C_(2) (""^(n+1)C_(3)+""^(n+1)C_(4) + ""^(n+1)C_(5)+ ...+ ""^(n+1)C_(n+1))`
` + ...+ ""^(n)C_(n-1)(""^(n +1)C_(n) + ""^(n+1)C_(n+1))`
`= ""^(n+1)C_(1)*""^(n)C_(0) + ""^(n+1)C_(2) (""^(n)C_(0) + ""^(n)C_(1)) + ""^(n+1)C_(3) (""^(n)C_(0) + ""^(n)C_(1) + ""^(n)C_(2))`
` + ...+ ""^(n+1)C_(n+1) (""^(n)C_(0) + ""^(n)C_(1) + ""^(n)C_(2)+ ...+ ""^(n)C_(n-1))`
`(""^(n)C_(0) + ""^(n)C_(1)) *""^(n)C_(0)+ (""^(n)C_(1) + ""^(n)C_()) ( ""^(n)C_(2)+ ""^(n)C_(1))+ (""^(n)C_(2) + ""^(n)C_(3)) (""^(n)C_(0) + ""^(n)C_(1) + ""^(n)C_(2))`
`+ ...+ (""^(n)C_(n) + ""^(n)C_(n-1))(""^(n)C_(0) + ""^(n)C_(1)+ ""^(n)C_(2) + ...+ ""^(n)C_(n-1)) + n`
`(""^(n)C_(0))^(2) + (""^(n)C_(1))^(2) + (""^(n)C_(2))^(2) + ...+ (""^(n)C_(n-1))^(2)`
` + {""^(n)C_(0) *""^(n)C_(1) + ""^(n)C_(0) * ""^(n)C_(2) + ""^(n)C_(0) *""^(n)C_(3)`
`+...+""^(n)C_(0) * ""^(n)C_(n-1) + ...+ ""^(n)C_(n-2) + ""^(n)C_(n-1)} + 2^(n) - 1 + n`
` (""^(n)C_(0) + ""^(n)C_(1) + ""^(n)C_(2) + ...+ ""^(n)C_(n-1))^(2) + 2^(n) - 1 + n`
`= (2^(n) -1)^(2) + 2^(n) -1 + n = 2 ^(2N) - 2^(n) + n`


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