题目内容

19.$C_{27}^1+C_{27}^2+C_{27}^3+…+C_{27}^{27}$除以9的余数为(  )
A.2B.4C.7D.8

分析 把所给的式子化为 (9-1)9-1,即 ${C}_{9}^{0}$•99-${C}_{9}^{1}$•98+${C}_{9}^{2}$•97-${C}_{9}^{3}$96+…+${C}_{9}^{8}$•9-${C}_{9}^{9}$-1,由此求得该式除以9的余数.

解答 解:$C_{27}^1+C_{27}^2+C_{27}^3+…+C_{27}^{27}$=${C}_{27}^{0}$+$C_{27}^1+C_{27}^2+C_{27}^3+…+C_{27}^{27}$-1
=(1+1)27-1=(9-1)9-1
=${C}_{9}^{0}$•99-${C}_{9}^{1}$•98+${C}_{9}^{2}$•97-${C}_{9}^{3}$96+…+${C}_{9}^{8}$•9-${C}_{9}^{9}$-1,
显然,前9项都能被9整除,故该式除以9的余数为-${C}_{9}^{9}$-1=-2,
即该式除以9的余数为7,
故选:C.

点评 本题主要考查二项式定理的应用,二项式展开式的通项公式,属于基础题.

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