Question

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

Answer 1

Correct Answer - `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)`