1
Question
If n≥r−1 and
nCr−1+n+1Cr−1+n+2Cr−1+....+2nCr−1=2n+1Cr2−182−nCr
then the least possible value of n is
If n≥r−1 and
nCr−1+n+1Cr−1+n+2Cr−1+....+2nCr−1=2n+1Cr2−182−nCr
then the least possible value of n is
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Solution
The correct option is D 13
nCr−1+n+1Cr−1+n+2Cr−1+n+3Cr−1+...2nCr−1
=nCr+nCr−1+n+1Cr−1+n+2Cr−1+n+3Cr−1+...2nCr−1−nCr
=n+1Cr+n+1Cr−1+n+2Cr−1+n+3Cr−1+...2nCr−1−nCr
:
:
2nCr−1+2nCr−nCr
=2n+1Cr−nCr
=2n+1Cr2−182−nCr
Hence
2n+1Cr=2n+1Cr2−182
Therefore
r=r2−182
r2−r−182=0
(r−14)(r+13)=0
Hence
r=14
Now
n≥r−1
Hence
n≥14−1
n≥13
nCr−1+n+1Cr−1+n+2Cr−1+n+3Cr−1+...2nCr−1
=nCr+nCr−1+n+1Cr−1+n+2Cr−1+n+3Cr−1+...2nCr−1−nCr
=n+1Cr+n+1Cr−1+n+2Cr−1+n+3Cr−1+...2nCr−1−nCr
:
:
2nCr−1+2nCr−nCr
=2n+1Cr−nCr
=2n+1Cr2−182−nCr
Hence
2n+1Cr=2n+1Cr2−182
Therefore
r=r2−182
r2−r−182=0
(r−14)(r+13)=0
Hence
r=14
Now
n≥r−1
Hence
n≥14−1
n≥13
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