For n >3 ,1.2nCr−2.3nCr−1+.....+(−1)r(r+1)(r+2) is
A
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B
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C
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D
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Solution
The correct option is B we have (1+x)n=nC0+nC1+nC2x2+...... +nCr−1xr−1+nCrxr+....+nCnxn....(1) and (1+x)−3=1−3C1x+4C2x2.....+(−1)r−1r+1Cr−1xr−1+(−1)r r+2Crxr+......to∞.....(2) Multiply (1) and (2), we get (1+x)n−3=(nC0+nC1x+nC2x2+....+nCnxn)(1−3C1x+4C2x2−...∞) Clearly , the coefficient of xr from the product in R.H.S is 1.nCr−3C1.nCr−1+4C2.nCr−2−....+(−1)rr+2Cr.nC0 =nCr−3nC1.nCr−1+4.32.1nCr−2−5.42.1nCr−3 +...+(1)r(r+2)(r+1)2.1 =12[1.2nCr−2.3nCr−1+4.3nCr−2−....+(−1)r(r+2)(r+1)] ∴ Required series =2× coefficient of xr in (1+x)n−3=2.n−3Cr