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Question

2C0+22C12+23C23+24C34+.........+2n+1Cnn+1=3n+11n+1

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Solution

2C0+22C12+23C23+24C34+.........+2n+1Cnn+1=3n+11n+1

=nr=0nCrr+12r1

=1n+1 nr=0n+1r+1nCr2r+1

=1n+1 nr=0n+1Cr+12r+1

=1n+1[n+1C12+n+1C222+...n+1Cn+12n+1+n+1C020n+1C020]

=1n+1[(1+2)n+11]=3n+11n+1

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