If n≥2 then (a−1).C1−(a−2).C2+(a−3).C3−……(−1)n−1(a−n).Cn=
A
0
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B
a−1
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C
a
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D
a+1
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Solution
The correct option is Ba (a−1)C1−(a−2)C2+(a−3)C3−...(−1)n(a−n)Cn =[aC1−aC2+aC3...]−[C1−2C2+C3...] =a(C1−C2+C3...)−(C1−2C2+C3...) =a(1+(−1+C1−C2+C3...))−(C1−2C2+C3...) =a+a(−1+C1−C2+C3...)−(C1−2C2+C3...) =a+−a(1−x)n|x=1−n(1−x)n−1|x=1 =a