KCET · Maths · Binomial Theorem
The value of \({ }^{10} \mathrm{C}_{1}+{ }^{10} \mathrm{C}_{2}+{ }^{10} \mathrm{C}_{3}+\ldots+{ }^{10} \mathrm{C}_{9}\) is
- A \(2^{10}\)
- B \(2^{11}\)
- C \(2^{10}-2\)
- D \(2^{10}-1\)
Answer & Solution
Correct Answer
(C) \(2^{10}-2\)
Step-by-step Solution
Detailed explanation
Since, \((1+x)^{n}={ }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2}\)
\(+\ldots+{ }^{n} C_{n} x^{n}\)
Put \(x=1\) and \(n=10\), we get
\[
\begin{aligned}
&2^{10}={ }^{10} \mathrm{C}_{0}+{ }^{10} \mathrm{C}_{1}+{ }^{10} \mathrm{C}_{2}+\ldots+{ }^{10} \mathrm{C}_{10} \\
&2^{10}=1+{ }^{10} \mathrm{C}_{1}+{ }^{10} \mathrm{C}_{2}+\ldots+{ }^{10} \mathrm{C}_{9}+1 \\
&2^{10}-2={ }^{10} \mathrm{C}_{1}+{ }^{10} \mathrm{C}_{2}+\ldots+{ }^{10} \mathrm{C}_{9}
\end{aligned}
\]
\(+\ldots+{ }^{n} C_{n} x^{n}\)
Put \(x=1\) and \(n=10\), we get
\[
\begin{aligned}
&2^{10}={ }^{10} \mathrm{C}_{0}+{ }^{10} \mathrm{C}_{1}+{ }^{10} \mathrm{C}_{2}+\ldots+{ }^{10} \mathrm{C}_{10} \\
&2^{10}=1+{ }^{10} \mathrm{C}_{1}+{ }^{10} \mathrm{C}_{2}+\ldots+{ }^{10} \mathrm{C}_{9}+1 \\
&2^{10}-2={ }^{10} \mathrm{C}_{1}+{ }^{10} \mathrm{C}_{2}+\ldots+{ }^{10} \mathrm{C}_{9}
\end{aligned}
\]
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