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Question

If an=nr=01nCr, then the value of nr=0n2rnCr is

A
n2an
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B
14an
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C
nan
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D
0
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Solution

The correct option is C 0
an=nr=01nCr,nr=0n2rnCr=?
an=1nC0+1nC1+1nC2+......+1nCn1+1nCn
=1nC0+1nC1+........+1nC1+1nC0
Because nCr=nCnr
an=1+[2nC1+2nC2+.....]+1 (nC0=nCn=1)
an22=[1nC1+1nC2+....]
rn=0rnCr=0+1nC1+2nC2+.....+n1nCn1+nnCn
=1nC1+2nC2+.....+n2nC2+n1nC1+n
=[1+(n1)nC1+2+(n2)nC2+....]+n
=n[1nC1+1nC2+....]+n
=n(an22)+n
nr=0rnCr=n(an22+1)=n(an2)
nr=0n2rnCr=nr=0nnCr2nr=0rnCr
=n(an)2(nan2)
=nannan
=0.
Hence, the answer is 0.

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