2
Question
The sum 1⋅ 20C1−2⋅ 20C2+3⋅ 20C3−⋯−20⋅ 20C20 is equal to
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Solution
Using r⋅nCr=n⋅ n−1Cr−1
∴20∑r=1(−1)r−1⋅r⋅ 20Cr=20∑r=1(−1)r−1⋅20⋅ 19Cr−1
=20(19C0− 19C1+ 19C2−⋯− 19C19)
=20×0=0
∴20∑r=1(−1)r−1⋅r⋅ 20Cr=20∑r=1(−1)r−1⋅20⋅ 19Cr−1
=20(19C0− 19C1+ 19C2−⋯− 19C19)
=20×0=0
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