If C0,C1,C2,⋯Cn, are the binomial coefficients of the expansion (1+x)n, where n is even, then C0+(C0+C1)+(C0+C1+C2)+⋯⋯+(C0+C1+C2+⋯+Cn−1)=
A
n⋅2n−1
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B
n⋅2n−1−n
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C
n⋅2n+n
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D
(n−1)⋅2n
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Solution
The correct option is An⋅2n−1 Given expansion is (1+x)n
Let S=C0+(C0+C1)+(C0+C1+C2)+⋯⋯+(C0+C1+C2+⋯+Cn−1)⇒S=n⋅C0+(n−1)C1+(n−2)C2+⋯+Cn−1⇒S=n−1∑r=0(n−r)nCr⇒S=n−1∑r=0nnCr−n−1∑r=0rnCr⇒S=n[2n−nCn]−nn−1∑r=1n−1Cr−1⇒S=n[2n−nCn]−n[2n−1−1]∴S=n⋅2n−1