Sum of the series 3C1−4C2+5C3−6C4+⋯ upto n terms is (where Cr=nCr)
A
−1
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B
2
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C
−2
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D
1
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Solution
The correct option is B2 Let S=3C1−4C2+5C3−7C5+⋅⋅⋅ upto n terms =n∑r=1(−1)r−1(r+2)Cr =n∑r=1(−1)r−1⋅r⋅nCr+2n∑r=1(−1)r−1⋅nCr =n∑r=1(−1)r−1⋅n⋅n−1Cr−1−2n∑r=1(−1)r⋅nCr =n⋅0−2n∑r=0nCr(−1)r+2nC0(∵n∑r=0nCr(a)r=(1+a)n) =2