有道数学题不会,求解
已知字母a.b满足根号a-1+根号b-2=0 求ab分之1+(a+1)(b+1)分之1+(a+2)(b+2)分之1......(a+2008)(b+1008)分之1
已知字母a.b满足根号a-1+根号b-2=0 求ab分之1+(a+1)(b+1)分之1+(a+2)(b+2)分之1......(a+2008)(b+1008)分之1 最后一个应该是:(a+2008)(b+2008)分之1 字母a.b满足根号a-1+根号b-2=0, 因为算数平方根永远大于等于0, 因此:a-1=0;b-2=0 所以:a=1;b=2 所以 原式=1/(1*2)+1/(2*3)+1/(3*4)+...+1/(2009*2010) =(2-1)/(1*2)+(3-2)/(2*3)+...+(2010-2009)/(2010*2009) =1-1/2+1/2-1/3+1/3-1/4+...+1/2009-1/2010 =1-1/2010 =2009/2010 记住这个公式:1/[n*(n+k)]=[1/n-1/(n+k)]/k n和k都是自然数
∵ √(a-1)+√(b-2)=0, ∴ a-1=0且b-2=0,即a=1,b=2. 原式=[1/(1×2)]+[1/(2×3)]+...+[1/(2009×2010)] =1-(1/2)+(1/2)-(1/3)+...-(1/2009)+(1/2009)-(1/2010) =1-(1/2010) =2009/2010.
解:√(a-1)+√(b-2)=0 ∵√(a-1)≥0,√(b-2)≥0 ∴√(a-1)=0,得:a=1 √(b-2)=,得b=2 那么:ab分之1+(a+1)(b+1)分之1+(a+2)(b+2)分之1......(a+2008)(b+2008)分之1 =1/(1×2)+1/(2×3)+1/(3×4)+……+1/(2009×2010) =1-1/2+1/2-1/3+1/3-1/4+……+1/2008-1/2009+1/2009-1/2010 =1-1/2010 =2009/2010。
sqrt(a-1)+sqrt(b-2)=0: a=1, b=2 原式=1/(1*2)+1/(2*3)+1/(3*4)+...+1/(2009*2010) =(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2009-1/2010) =1-1/2010 =2009/2010
问:数列问题已知数列{an}满足a1=根号3,a(n+1)=an*2^n 求an
答:a(n+1)=an×2^n, a2/a1=2 a3/a2=2² ………… an/a(n-1)=2^(n-1), ∴迭乘,得an/a1=2×2²...详情>>
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