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Question

# The sum of the series n∑r=1(−1)r−1.nCr(a−r) is

A
a
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B
0
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C
n.2n1+a
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D
None of these
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Solution

## The correct option is A a∑nr=1(−1)r−1ncr(a−r)Putn=1⇒nc1(a−1)−nc2(a−2)+nc3(a−1)+.......(−1)n−1ncn(a−n)⇒a[nc1−nc2+nc3−nc4+nc5+.......−ncn]−nc1+2nc2−3nc3+4nc4.........=a[nc1−nc2+nc3−nc4+nc5+......]−[nc1−2nc2+3nc3+4nc4+.......](1−x)n=nc0−nc1x+nc2x2−nc3x3.......[Accordingtothisformula]⇒Add&subtractnc0inahaveequation&takingoutcommon(−1)also.⇒a[−(nc0−nc1+nc2−nc3+......)+nco]−[nc1−2nc2+3nc3−4nc4]Putx=1(1−1)n=nc0−nc1+nc2−nc3=0So,a[−0+nc0]−[nc1−2nc2+3nc3−4nc4+.........]→(i)(1+n)n=nc0+nc1x+nc2x2+nc3x3+nc4x4+.......[←formula]Diff.w.r.t.n,wegetn(1+x)n−1=nc1+2xnc2+3xnc3+4xnc4+....Putx=−1So,x(1−x)n−1=nc1−2xnc2+3x2nc3+4nc4+.........0=nc1−2nc2+3nc3−4nc4So, put the value in equation in (1)=a[nc0]−[0]⇒a(1)=a∑nr=1(−1)r−1ncr(a−r)=a

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