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101c0+101c2+101c3....101c50 help me find the sum in binomial expansion

Prasi , 9 Years ago
Grade 12
anser 1 Answers
Sai Pranav
We know that
101C0+101C1+101C2+101C3+........+101C50+101C51+101C52+........+101C100+101C101=2101
But 
nCr=nCn-r
Therefore
101C51=101C50
101C52=101C49
.….............
101C100=101C1
 
101C101=101C0
Hence
101C0+101C1+101C2+101C3+........+101C50+101C50+.......+101C1+101C0=2101
Which is
2(101C0+101C1+101C2+101C3+........+101C50)=2101
101C0+101C1+101C2+101C3+........+101C50=\frac{2^{101}}{2}
THUS
101C0+101C1+101C2+101C3+........+101C50=2100
 
 
Last Activity: 9 Years ago
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