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QJEEAdv10Ma106d: JEE Advanced 2010 Mathematics - Class 11 JEEAdv10Ma106d Solution: For let and denote, respectively, the coefficient of in the…

For r=0, 1,....,r = 0,\,1,...., let Ar, Br{A_r},\,{B_r} and Cr{C_r} denote, respectively, the coefficient of Xr{X^r} in the expansions of (1+x)10,{\left( {1 + x} \right)^{10}}, (1+x)20{\left( {1 + x} \right)^{20}} and (1+x)30.{\left( {1 + x} \right)^{30}}.
Then ∑r=110Ar(B10Br−C10Ar)\sum\limits_{r = 1}^{10} {{A_r}\left( {{B_{10}}{B_r} - {C_{10}}{A_r}} \right)} is equal to

Multiple Choice Options

  • A (B10−C10)\left( {{B_{10}} - {C_{10}}} \right)
  • B A10(B210C10A10){A_{10}}\left( {{B^2}_{10}{C_{10}}{A_{10}}} \right)
  • C 00
  • D C10−B10{{C_{10}} - {B_{10}}}

Step-by-Step Verified Solution

Ar = Coefficient of xr in (1+x)10=10Cr{(1 + x)^{10}} = {}^{10}{C_r}

Br = Coefficient of xr in (1+x)20=20Cr{(1 + x)^{20}} = {}^{20}{C_r}

Cr = Coefficient of xr in (1+x)30=30Cr{(1 + x)^{30}} = {}^{30}{C_r}

∴\therefore ∑r=110Ar(B10Br−C10Ar)\sum\limits_{r = 1}^{10} {{A_r}({B_{10}}{B_r} - {C_{10}}{A_r})}

=∑r=110ArB10Br−∑r=110ArC10Ar= \sum\limits_{r = 1}^{10} {{A_r}{B_{10}}{B_r} - \sum\limits_{r = 1}^{10} {{A_r}{C_{10}}{A_r}} }

=∑r=11010Cr 20C10 20Cr−∑r=11010Cr 30C10 10Cr= \sum\limits_{r = 1}^{10} {{}^{10}{C_r}\,{}^{20}{C_{10}}\,{}^{20}{C_r} - \sum\limits_{r = 1}^{10} {{}^{10}{C_r}\,{}^{30}{C_{10}}\,{}^{10}{C_r}} }

=∑r=11010C10−r 20C10 20Cr−∑r=11010C10−r 30C10 10Cr= \sum\limits_{r = 1}^{10} {{}^{10}{C_{10 - r}}\,{}^{20}{C_{10}}\,{}^{20}{C_r} - \sum\limits_{r = 1}^{10} {{}^{10}{C_{10 - r}}\,{}^{30}{C_{10}}\,{}^{10}{C_r}} }

=20C10∑r=11010C10−r .  20Cr−30C10∑r=11010C10−r .  10Cr= {}^{20}{C_{10}}\sum\limits_{r = 1}^{10} {{}^{10}{C_{10 - r}}\,.\,\,{}^{20}{C_r} - {}^{30}{C_{10}}\sum\limits_{r = 1}^{10} {{}^{10}{C_{10 - r}}\,.\,\,{}^{10}{C_r}} }

=20C10(30C10−1)−30C10(20C10−1)= {}^{20}{C_{10}}\left( {{}^{30}{C_{10}} - 1} \right) - {}^{30}{C_{10}}\left( {{}^{20}{C_{10}} - 1} \right)

=30C10−20C10=C10−B10= {}^{30}{C_{10}} - {}^{20}{C_{10}} = {C_{10}} - {B_{10}}

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