求:小学奥数题1+11+111+1111+11111+……+100个1=?
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1+11+111+1111+11111+……+ n个1= n+10(n-1)+100(n-2)+...+10^(n-2)*2+10^(n-1)
=n+10n+100n+1000n+...+10^(n-2)*n+10^(n-1)*n-10-200-3000-...-(n-2)10^(n-2)-(n-1)10^(n-1)
=n(1-10^n)/(1-10)-(10+100+1000+...+10^(n-1))-(100+1000+...+10^(n-1))-(1000+...+10^(n-1))-10^(n-1)
=n(10^n-1)/9-10[10^(n-1)-1]/9-100[10^(n-2)-1]/9-...-10^(n-1)[10^(n-n+1)-1]/9
=[n(10^n-1)-10^n+10-10^n+100-...-10^n+10^(n-1)]/9
=[n(10^n-1)-(n-1)10^n+(10^n-10)/9]/9
=[-n+10^n+(10^n-10)/9]/9
=(-9n+9*10^n+10^n-10)/81
=[10^(n+1)-9n-10]/81
=n+10n+100n+1000n+...+10^(n-2)*n+10^(n-1)*n-10-200-3000-...-(n-2)10^(n-2)-(n-1)10^(n-1)
=n(1-10^n)/(1-10)-(10+100+1000+...+10^(n-1))-(100+1000+...+10^(n-1))-(1000+...+10^(n-1))-10^(n-1)
=n(10^n-1)/9-10[10^(n-1)-1]/9-100[10^(n-2)-1]/9-...-10^(n-1)[10^(n-n+1)-1]/9
=[n(10^n-1)-10^n+10-10^n+100-...-10^n+10^(n-1)]/9
=[n(10^n-1)-(n-1)10^n+(10^n-10)/9]/9
=[-n+10^n+(10^n-10)/9]/9
=(-9n+9*10^n+10^n-10)/81
=[10^(n+1)-9n-10]/81
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