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Question

If (1−x)−n=a0+a1x+a2x2+........+arxr+....., then a0+a1+a2+......+ar is equal to

A
n(n+1)(n+2)....(n+r)r!
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B
(n+1)(n+2)....(n+r)r!
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C
n(n+1)(n+2)....(n+r1)r!
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D
None of the above
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Solution

The correct option is B (n+1)(n+2)....(n+r)r!
we have,
(1x)n=a0+a1x+a2x2+........+arxr+.....
and
(1x)1=1+x+x2+x3+....+xr+.....
Hence,
a0+a1+a2+......+ar
= coefficient of xr in the product of two series
= coefficient of xr in (1x)n(1x)1
= coefficient of xr in (1x)(n+1)
=(n+1)(n+2)....(n+r)r!

Alternate Solution:
General term in the expansion of (1x)n will be
Tr+1=n+r1Crxr
hence a0+a1+a2+......+ar
=n1C0+nC1+n+1C2++n+r1Cr
Now as we know that
n1C0+nC1=nC0+nC1=n+1C1
similarly n+1C1+n+1C2=n+2C2
Hence
n1C0+nC1+n+1C2++n+r1Cr=n+rCr=(n+r)!n!r!=(n+r)(n+r1)(n+2)(n+1)r!

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