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Find the remainder when (sum_(r=1)^(5)""^(20)C_(2r-1))^(6) is divided by 11. |
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Answer» 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. |
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