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Evaluate sum(i=0)^(n)sum(j=0)^(n) ""^(n)...

Evaluate `sum_(i=0)^(n)sum_(j=0)^(n) ""^(n)C_(j) *""^(j)C_(i), ile j ` .

Text Solution

Verified by Experts

We have , `sum_(i=0)^(n)sum_(j=0)^(n) ""^(n)C_(j) *""^(j)C_(i)`
`= ""^(n)C_(1) (""^(1)C_(0) + ""^(1)C_(1)) +""^(n)C_(2) (""^(2)C_(0) + ""^(2)C_(1) + ""^(2)C_(2))`
` + ""^(n)C_(3) (""^(3)C_(0) + ""^(3)C_(1) + ""^(3)C_(2) + ""^(3)C_(3))`
` + ""^(n)C_(4) (""^(4)C_(0) + ""^(4)C_(1) + ""^(4)C_(2) + ""^(4)C_(3)+""^(4)C_(4))`
`+ ...+ ""^(n)C_(n)(""^(n)C_(0) + ""^(n)C_(1) + ""^(n)C_(2) + ...+ ""^(n)C_(n))`
`=""^(n)C_(1)(2)+""^(n)C_(2)(2)^(2) + ""^(n)C_(3)(2)^(3) + ... + ""^(n)C_(n)(2)^(n) = (1+ 2)^(n) - 1 `
`= 3^(n) -1`
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