For $r = 0,\$…,10 let $A_{r},\ B_{r}\ and\ C_{r}$ denotes, respectively, the coefficient of $x^{r}$ in the ${(1 + x)}^{10},\ {(1 + x)}^{20},\$ and ${(1 + x)}^{30}$. Then $$$\sum_{r = 1}^{10}{A_{r}(B_{10}B_{r} - C_{10}A_{r})\ }$ is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
$A_{r} =$ Coefficient of $x^{r}$ in ${(1 + x)}^{10} = \ ^{10}C_{r}$
$B_{r} =$ Coefficient of $x^{r}$ in ${(1 + x)}^{20} = \ ^{20}C_{r}$
$C_{r} =$ Coefficient of $x^{r}$ in ${(1 + x)}^{30} = \ ^{30}C_{r}$
$	herefore\sum_{r = 1}^{10}\ A_{r}\left( B_{10}B_{r} - C_{10}A_{r} ight) = \sum_{r = 1}^{10}\ A_{r}B_{10}B_{r}, - \sum_{r = 1}^{10}{A_{r}C_{10}A_{r}\ }$
$= \sum_{r = 1}^{10}{\ ^{10}C_{r}\ ^{20}C_{10}\ ^{20}C_{r}\ }\sum_{r = 1}^{10}{\ ^{10}C_{r}\ ^{30}C_{10}\ ^{10}C_{r}l\ }$
$\sum_{r = 1}^{10}{\ ^{10}C_{10 - rl}\ ^{20}C_{10}\ ^{20}C_{r} - \ }\sum_{r = 1}^{10}\ \ ^{10}C_{10 - r}\ ^{30}C_{10}\ ^{10}C_{r}l$
$= \ ^{20}C_{10}\sum_{r = 1}^{10}{\ ^{10}C_{10 - r}\ ^{20}C_{r} - \ ^{30}C_{10}\ }\sum_{r = 1}^{10}\ \ ^{10}C_{10 - r}\ ^{10}C_{r}$
$= \ ^{20}C_{10}\left( \ ^{30}C_{10} - 1 ight) - \ ^{30}C_{10}(\ ^{20}C_{10} - 1)$
${= \ }^{20}C_{10}\left( \ ^{30}C_{10} - 1 ight) - \ ^{30}C_{10}\left( \ ^{20}C_{10} - 1 ight)$
$= \ ^{30}C_{10} - \ ^{20}C_{10} = C_{10} - B_{10}$

Q. For $r = 0,\$…,10 let $A_{r},\ B_{r}\ and\ C_{r}$ denotes, respectively, the coefficient of $x^{r}$ in the ${(1 + x)}^{10},\ {(1 + x)}^{20},\$ and ${(1 + x)}^{30}$. Then $$$\sum_{r = 1}^{10}{A_{r}(B_{10}B_{r} - C_{10}A_{r})\ }$ is equal to

Explanation

Reviewed by Dr. Abubacker

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