JEE Advanced · Mathematics · 6. Binomial Theorem
For \(r=0,1, \ldots, 10\), let \(A_r, B_r\) 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 \(C_{10}-B_{10}\)
Answer & Solution
Correct Answer
(D) \(C_{10}-B_{10}\)
Step-by-step Solution
Detailed explanation
\(A_r=\) Coefficient of \(x^r\) in
\((1+x)^{10}={ }^{10} C_r \)
\( B_r=\text { Coefficient of } x^r \text { in } \)
\( (1+x)^{20}={ }^{20} C_r \)
\( C_r=\text { Coefficient of } x^r \text { in } \)
\( (1+x))^{30}={ }^{30} C_r \)
\( \therefore \sum_{r=1}^{10} A_r\left(B_{10} B_r-C_{10} A_r\right) \)
\( =\sum_{r=1}^{10} A_r B_{10} B_r-\sum_{r=1}^{10} A_r C_{10} A_r \)
\( =\sum_{r=1}^{10}{ }^{10} C_r{ }^{20} C_{10}{ }^{20} C_r \)
\( -\sum_{r=1}^{10}{ }^{10} C_r{ }^{30} C_{10}{ }^{10} C_r \)
\( =\sum_{r=1}^{10}{ }^{10} C_{10-r}{ }^{20} C_{10}{ }^{20} C_r \)
\( -\sum_{r=1}^{10}{ }^{10} C_{10-r}{ }^{30} C_{10}{ }^{10} C_r \)
\(={ }^{20} C_{10} \sum_{r=1}^{10}{ }^{10} C_{10-r} \cdot{ }^{20} C_r \)
\( -{ }^{30} C_{10} \sum_{r=1}^{10}{ }^{10} C_{10-r}{ }^{10} C_r \)
\( ={ }^{20} C_{10}\left({ }^{30} C_{10}-1\right)-{ }^{30} C_{10}\left({ }^{20} C_{10}-1\right) \)
\( ={ }^{30} C_{10}-{ }^{20} C_{10}=C_{10}-B_{10}\)
\((1+x)^{10}={ }^{10} C_r \)
\( B_r=\text { Coefficient of } x^r \text { in } \)
\( (1+x)^{20}={ }^{20} C_r \)
\( C_r=\text { Coefficient of } x^r \text { in } \)
\( (1+x))^{30}={ }^{30} C_r \)
\( \therefore \sum_{r=1}^{10} A_r\left(B_{10} B_r-C_{10} A_r\right) \)
\( =\sum_{r=1}^{10} A_r B_{10} B_r-\sum_{r=1}^{10} A_r C_{10} A_r \)
\( =\sum_{r=1}^{10}{ }^{10} C_r{ }^{20} C_{10}{ }^{20} C_r \)
\( -\sum_{r=1}^{10}{ }^{10} C_r{ }^{30} C_{10}{ }^{10} C_r \)
\( =\sum_{r=1}^{10}{ }^{10} C_{10-r}{ }^{20} C_{10}{ }^{20} C_r \)
\( -\sum_{r=1}^{10}{ }^{10} C_{10-r}{ }^{30} C_{10}{ }^{10} C_r \)
\(={ }^{20} C_{10} \sum_{r=1}^{10}{ }^{10} C_{10-r} \cdot{ }^{20} C_r \)
\( -{ }^{30} C_{10} \sum_{r=1}^{10}{ }^{10} C_{10-r}{ }^{10} C_r \)
\( ={ }^{20} C_{10}\left({ }^{30} C_{10}-1\right)-{ }^{30} C_{10}\left({ }^{20} C_{10}-1\right) \)
\( ={ }^{30} C_{10}-{ }^{20} C_{10}=C_{10}-B_{10}\)
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