已知a,bεR+，nεN, 求证（a+b）（a^n+b^n)≤2(a^(n+1)+b^(n+1))
热心网友
a,bεR+，nεN == (a-b)(a^n - b^n)= (a-b)^2 *[a^(n-1)+b*a^(n-2)+...+a*b^(n-2)+b^(n-1)] = 0== a^(n+1) +b^(n+1) = b*a^n +a*b^n== 2*(a^(n+1)+b^(n+1)) = b*a^n +a*b^n + a^(n+1)+b^(n+1)= (a+b)(a^n+b^n)
已知a,bεR+，nεN, 求证（a+b）（a^n+b^n)≤2(a^(n+1)+b^(n+1))
a,bεR+，nεN == (a-b)(a^n - b^n)= (a-b)^2 *[a^(n-1)+b*a^(n-2)+...+a*b^(n-2)+b^(n-1)] = 0== a^(n+1) +b^(n+1) = b*a^n +a*b^n== 2*(a^(n+1)+b^(n+1)) = b*a^n +a*b^n + a^(n+1)+b^(n+1)= (a+b)(a^n+b^n)