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JEE Advanced 2010Paper 2
Exam
JEE Advanced
Year
2010
Paper
Paper 2
Subject
Mathematics
Chapter
Binomial Theorem

JEE Advanced 2010 · Paper 2 · Mathematics · Binomial Theorem

For r=0,1, , 10 , let A_r, B_r and C_r denote, respectively, the...

For r=0,1,…,10r=0,1, \ldots, 10, let Ar,BrA_r, B_r and CrC_r denote, respectively, the coefficient of xrx^r in the expansions of (1+x)10(1+x)^{10}, (1+x)20and(1+x)^{20} and(1+x)^{30}.Then. Then\sum_{r=1}^{10} A_r(B10Br−C10Ar)\left(B_{10} B_r-C_{10} A_r\right) is equal to

Options

  1. AB10−C10B_{10}-C_{10}
  2. BA10(B102−C10A10)A_{10}\left(B_{10}^2-C_{10} A_{10}\right)
  3. C0
  4. DC10−B10C_{10}-B_{10}

Correct answer

D. C10−B10C_{10}-B_{10}

Step-by-step solution

A_r=CoefficientofCoefficient ofx^rinin(1+x)10=10CrBr= Coefficient of xr in (1+x)20=20CrCr= Coefficient of xr in (1+x))30=30Cr∴∑r=110Ar(B10Br−C10Ar)=∑r=110ArB10Br−∑r=110ArC10Ar=∑r=11010Cr(2010)20Cr−∑r=11010Cr(3010)10Cr=∑r=110(1010−r)(2010)20Cr−∑r=110(1010−r)(3010)10Cr\begin{aligned}& (1+x)^{10}={ }^{10} C_r \\ & B_r=\text{ Coefficient of }x^r \text{ in }\\ & (1+x)^{20}={ }^{20} C_r \\ & C_r=\text{ Coefficient of }x^r \text{ in }\\ & (1+x))^{30}={ }^{30} C_r \\ & \therefore \sum_{r=1}^{10} A_r\left(B_{10} B_r-C_{10} A_r\right) \\ & =\sum_{r=1}^{10} A_r B_{10} B_r-\sum_{r=1}^{10} A_r C_{10} A_r \\ & =\sum_{r=1}^{10}{ }^{10} C_r\binom{20}{10}{ }^{20} C_r \\ & -\sum_{r=1}^{10}{ }^{10} C_r\binom{30}{10}{ }^{10} C_r \\ & =\sum_{r=1}^{10}\binom{10}{10-r}\binom{20}{10}{ }^{20} C_r \\ & -\sum_{r=1}^{10}\binom{10}{10-r}\binom{30}{10}{ }^{10} C_r \\ & \end{aligned}$\begin{aligned}& =\binom{20}{10} \sum_{r=1}^{10}\binom{10}{10-r} \cdot{ }^{20}C_r \\ & \quad-\binom{30}{10} \sum_{r=1}^{10}\binom{10}{10-r}{ }^{10} C_r \\ & =\binom{20}{10}\left(\binom{30}{10}-1\right)-\binom{30}{10}\left(\binom{20}{10}-1\right) \\ & =\binom{30}{10}-\binom{20}{10}=C_{10}-B_{10} \end{aligned}
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