总述:本次共解1题。其中
☆算术计算1题
〖1/1算式〗
作业:求算式 C(3,0)*2^0+C(3,1)*2^1+C(3,2)*2^2+C(3,3)*2^3 的值.
题型:数学计算
解:
C(3,0)*2^0+C(3,1)*2^1+C(3,2)*2^2+C(3,3)*2^3
=1*2^0+C(3,1)*2^1+C(3,2)*2^2+C(3,3)*2^3
=1*2^0+3*2^1+C(3,2)*2^2+C(3,3)*2^3
=1*2^0+3*2^1+3*2^2+C(3,3)*2^3
=1*2^0+3*2^1+3*2^2+1*2^3
=1*1+3*2^1+3*2^2+1*2^3
=1*1+3*2+3*2^2+1*2^3
=1*1+3*2+3*4+1*2^3
=1*1+3*2+3*4+1*8
=1+3*2+3*4+1*8
=1+6+3*4+1*8
=1+6+12+1*8
=1+6+12+8
=7+12+8
=19+8
=27 答案:C(3,0)*2^0+C(3,1)*2^1+C(3,2)*2^2+C(3,3)*2^3=27你的问题在这里没有得到解决?请到 热门难题 里面看看吧!