Binomial Theorem 2 Question 4
4.
For $r=0,1, \ldots, 10$, if $A_{r}, B_{r}$ and $C_{r}$ denote respectively the coefficient of $x^{r}$ in the expansions of $(1+x)^{10},(1+x)^{20}$ and $(1+x)^{30}$. Then, $\sum_{r=1}^{10} A_{r}\left(B_{10} B_{r}-C_{10} A_{r}\right)$ is equal to
(a) $B_{10}-C_{10}$
(b) $A_{10}\left(B_{10}^{2}-C_{10} A_{10}\right)$
(c) 0
(d) $C_{10}-B_{10}$
(2010)
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Answer:
Correct Answer: 4. (d)
Solution:
- $A_{r}=$ 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 $(1+x)^{30}={ }^{30} C_{r}$
$\therefore \sum_{r=1}^{10} A_{r}\left(B_{10} B_{r}-C_{10} A_{r}\right)=\sum_{r=1}^{10} A_{r} B_{10} B_{r}-\sum_{r=1}^{10} A_{r} C_{10} A_{r}$
$=\sum_{r=1}^{10}{ }^{10} C_{r}{ }^{20} C_{10}{ }^{20} C_{r}-\sum_{r=1}^{10}{ }^{10} C_{r}{ }^{30} C_{10}{ }^{10} C_{r}$
$=\sum_{r=1}^{10}{ }^{10} C_{10-r}{ }^{20} C_{10}{ }^{20} C_{r}-\sum_{r=1}^{10}{ }^{10} C_{10-r}{ }^{30} C_{10}{ }^{10} C_{r}$
$={ }^{20} C_{10} \sum_{r=1}^{10}{ }^{10} C_{10-r} \cdot{ }^{20} C_{r}-{ }^{30} C_{10} \sum_{r=1}^{10}{ }^{10} C_{10-r}{ }^{10} C_{r}$
$={ }^{20} C_{10}\left({ }^{30} C_{10}-1\right)-{ }^{30} C_{10}\left({ }^{20} C_{10}-1\right)$
$={ }^{30} C_{10}-{ }^{20} C_{10}=C_{10}-B_{10}$
