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Question

# If n is an integer greater than 1, then aâˆ’nC1(aâˆ’1)+nC2(aâˆ’2)âˆ’....+(âˆ’1)n(aâˆ’n)=

A
a
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B
0
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C
a2
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D
2n
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Solution

## The correct option is B 0a−nC1(a−1)+nC2(a−2)−...….+(−1)n(a−n)=n∑r=0(−1)rnCr(a−r)=n∑r=0(−1)rnCr⋅a−n∑r=0(−1)rnCr⋅r=an∑r=0(−1)rnCr−n∑r=1(−1)rnCr⋅r (when r=0, we get 0)=a(1−1)n−n∑r=1(−1)r⋅nrn−1Cr−1⋅r(nCr=nrn−1Cr−1) ((1+x)n=n∑r=0nCrxr when x=−1 (1−1)r=n∑r=0nCr(−1)r)=a(0)−n∑r=1(−1)rn⋅n−1Cr−1=0−nn∑r=1(−1)rn−1Cr−1=−n[−n−1C0+n−1C1−n−1C2………(−1)n−1n−1Cn−1]=−n[0]=0 [(1−1)n−1=n−1C0−n−1C1+n+1C2−n−1C3+.....=(−1)nn−1Cn−1=0].

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