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

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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