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JEE Advance - Mathematics (2010 - Paper 2 Offline - No. 6)

For $$r = 0,\,1,....,$$ let $${A_r},\,{B_r}$$ and $${C_r}$$ denote, respectively, the coefficient of $${X^r}$$ in the expansions of $${\left( {1 + x} \right)^{10}},$$ $${\left( {1 + x} \right)^{20}}$$ and $${\left( {1 + x} \right)^{30}}.$$
Then $$\sum\limits_{r = 1}^{10} {{A_r}\left( {{B_{10}}{B_r} - {C_{10}}{A_r}} \right)} $$ is equal to
$$\left( {{B_{10}} - {C_{10}}} \right)$$
$${A_{10}}\left( {{B^2}_{10}{C_{10}}{A_{10}}} \right)$$
$$0$$
$${{C_{10}} - {B_{10}}}$$

Explanation

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

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

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

$$\therefore$$ $$\sum\limits_{r = 1}^{10} {{A_r}({B_{10}}{B_r} - {C_{10}}{A_r})} $$

$$ = \sum\limits_{r = 1}^{10} {{A_r}{B_{10}}{B_r} - \sum\limits_{r = 1}^{10} {{A_r}{C_{10}}{A_r}} } $$

$$ = \sum\limits_{r = 1}^{10} {{}^{10}{C_r}\,{}^{20}{C_{10}}\,{}^{20}{C_r} - \sum\limits_{r = 1}^{10} {{}^{10}{C_r}\,{}^{30}{C_{10}}\,{}^{10}{C_r}} } $$

$$ = \sum\limits_{r = 1}^{10} {{}^{10}{C_{10 - r}}\,{}^{20}{C_{10}}\,{}^{20}{C_r} - \sum\limits_{r = 1}^{10} {{}^{10}{C_{10 - r}}\,{}^{30}{C_{10}}\,{}^{10}{C_r}} } $$

$$ = {}^{20}{C_{10}}\sum\limits_{r = 1}^{10} {{}^{10}{C_{10 - r}}\,.\,\,{}^{20}{C_r} - {}^{30}{C_{10}}\sum\limits_{r = 1}^{10} {{}^{10}{C_{10 - r}}\,.\,\,{}^{10}{C_r}} } $$

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

$$ = {}^{30}{C_{10}} - {}^{20}{C_{10}} = {C_{10}} - {B_{10}}$$

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