We know that
101C0+101C1+101C2+101C3+........+101C50+101C51+101C52+........+101C100+101C101=2101
But
nCr=nCn-r
Therefore
101C
51=
101C
50101C52=101C49
101C100=101C1
101C101=101C0
Hence
101C0+101C1+101C2+101C3+........+101C50+101C50+.......+101C1+101C0=2101
Which is
2(101C0+101C1+101C2+101C3+........+101C50)=2101
101C
0+
101C
1+
101C
2+
101C
3+........+
101C
50=

THUS
101C0+101C1+101C2+101C3+........+101C50=2100