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Find the sum sum(j=0)^(n) (""^(4n+1)C(j)...

Find the sum `sum_(j=0)^(n) (""^(4n+1)C_(j)+""^(4n+1)C_(2n-j))`.

Text Solution

Verified by Experts

The correct Answer is:
`2^(4n)+.^(4n+1)C_(n)`
`underset(j=0)overset(n)sum(.^(4n+1)C_(j) + .^(4n+1)C_(2n-j))= (.^(4n+1)C_(0) + .^(4n+1)C_(1)+"....." .^(4n+1)C_(n))+(.^(4n+1)C_(2n)+.^(4n+1)C_(2n-1)+"....."+.^(4n+1)C_(n))`
`= (.^(4n+1)C_(0)+.^(4n+1)C_(1)+"...."+.^(4n+1)C_(2n))+.^(4n+1)C_(n)`
` = 2^(4n) + .^(4n+1)C_(n)`
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Statement -1: sum_(r=0)^(n) r(""^(n)C_(r))^(2) = n (""^(2n -1)C_(n-1)) Statement-2: sum_(r=0)^(n) (""^(n)C_(r))^(2)= ""^(2n)C_(n)

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