大神们求助，新人请教这个怎么使用循环语句简化啊。

这个怎么使用循环语句简化啊。还有怎么对R1每一列的最大值筛选并且建立矩阵呢。
R = {{0.86, 0.93, 0.90, 0.93, 0.72, 0.66, 0.91}, {0.83, 0.96, 0.85,
0.96, 0.76, 0.59, 0.90}, {0.81, 0.94, 0.86, 0.94, 0.75, 0.54,
0.85}}
R1[[1, 1]] = R[[1, 1]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 1]\)])\), \(2\)]\))]
R1[[2, 1]] = R[[2, 1]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 1]\)])\), \(2\)]\))]
R1[[3, 1]] = R[[3, 1]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 1]\)])\), \(2\)]\))]
R1[[1, 2]] = R[[1, 2]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 2]\)])\), \(2\)]\))]
R1[[2, 2]] = R[[2, 2]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 2]\)])\), \(2\)]\))]
R1[[3, 2]] = R[[3, 2]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 2]\)])\), \(2\)]\))]
R1[[1, 3]] = R[[1, 3]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 3]\)])\), \(2\)]\))]
R1[[2, 3]] = R[[2, 3]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 3]\)])\), \(2\)]\))]
R1[[3, 3]] = R[[3, 3]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 3]\)])\), \(2\)]\))]
R1[[1, 4]] = R[[1, 4]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 4]\)])\), \(2\)]\))]
R1[[2, 4]] = R[[2, 4]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 4]\)])\), \(2\)]\))]
R1[[3, 4]] = R[[3, 4]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 4]\)])\), \(2\)]\))]
R1[[1, 5]] = R[[1, 5]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 5]\)])\), \(2\)]\))]
R1[[2, 5]] = R[[2, 5]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 5]\)])\), \(2\)]\))]
R1[[3, 5]] = R[[3, 5]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 5]\)])\), \(2\)]\))]
R1[[1, 6]] = R[[1, 6]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 6]\)])\), \(2\)]\))]
R1[[2, 6]] = R[[2, 6]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 6]\)])\), \(2\)]\))]
R1[[3, 6]] = R[[3, 6]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 6]\)])\), \(2\)]\))]
R1[[1, 7]] = R[[1, 7]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 7]\)])\), \(2\)]\))]
R1[[2, 7]] = R[[2, 7]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 7]\)])\), \(2\)]\))]
R1[[3, 7]] = R[[3, 7]]/Sqrt[(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(3\)]
\*SuperscriptBox[\((R[\([k, 7]\)])\), \(2\)]\))]
R1 = {{R1[[1, 1]], R1[[1, 2]], R1[[1, 3]], R1[[1, 4]], R1[[1, 5]],
R1[[1, 6]], R1[[1, 7]]}, {R1[[2, 1]], R1[[2, 2]], R1[[2, 3]],
R1[[2, 4]], R1[[2, 5]], R1[[2, 6]], R1[[2, 7]]}, {R1[[3, 1]],
R1[[3, 2]], R1[[3, 3]], R1[[3, 4]], R1[[3, 5]], R1[[3, 6]],
R1[[3, 7]]}}
这个怎么使用循环语句简化啊。还有怎么对R1每一列的最大值筛选并且建立矩阵呢。