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"if "(1+x)^(n)=C(0)+C(1).x+C(2).x^(2)+C(...

`"if "(1+x)^(n)=C_(0)+C_(1).x+C_(2).x^(2)+C_(3).x^(3)+......+C_(n).x^(n),` then prove that
`C_(0)+2C_(1)+4C_(2)+6C_(3)+…….+2n.C_(n)=1+n.2^(n)`

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The correct Answer is:
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`L.H.S. =C_(0)+2C_(1)+4C_(2)+6C_(3)+……+2nC_(n)`
`=1+2[n+(2.n(n-1))/(|ul2)+(3n(n-1)(n-2))/(|ul2)+….n.1]`
`=1+2n[1+(n-1)+((n-1)(n-2))/(|ul2)+......+1]`
`=1+2n[.^(n+1)C_(0)+^(n-1)C_(1)+^(n-1)C_(2)+....+^(n-1)C_(n-1)]`
`1+2n.2^(n-1)`
`=1+n.2^(n)=R.H.S.` Hence Proved
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