Correct Answer - 3
We have `E = (underset(r=1)overset(5)sum.^(20)C_(2r-1))^(6)`
We know that `.^(20)C_(1)+.^(20)C_(3)+.^(20)C_(5)+.^(20)C_(7)+"......"+.^(20)C_(19)=2^(19)`
Now, `.^(20)C_(1)=.^(20)C_(19),.^(20)C_(3)=.^(20)C_(17)"....."`etc.
`:. 2(.^(20)C_(1)+.^(20)C_(3)+"...."+.^(20)C_(9))=2^(19)`
`rArr .^(20)C_(1)+.^(20)C_(3)+"....."+.^(20)C_(9)=2^(18)`
`rArr E = (2^(18))^(6) = 2^(108)`
`= 8(2^(5))^(21)`
`= 8(33-1)^(21)`
`= 8(33k-1)`
`= 8 xx 33k - 8`
`= 11(8xx3k-1)+3`
Therefore, the remainder when `(underset(r=1)overset(5)sum.^(20)C_(2r-1))^(6)` is divided by 11 is 3.
