WBJEE · Maths · Binomial Theorem
The value of the sum \(\left({ }^{n} C_{1}\right)^{2}+\left({ }^{n} C_{2}\right)^{2}+\left({ }^{n} C_{3}\right)^{2}+\ldots+\left({ }^{n} C_{n}\right)^{2}\) is
- A \(\left({ }^{2 n} C_{n}\right)^{2}\)
- B \({ }^{2 n} C_{n}\)
- C \({ }^{2 n} \mathrm{C}_{n}+1\)
- D \({ }^{2 n} C_{n}-1\)
Answer & Solution
Correct Answer
(D) \({ }^{2 n} C_{n}-1\)
Step-by-step Solution
Detailed explanation
We know that \[ \begin{array}{r} (1+x)^{n}={ }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2} \\ \quad+\ldots+{ }^{n} C_{n} x^{n} \\ \text { and }(x+1)^{n}={ }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1} \\ \quad+{ }^{n} C_{2} x^{n-2}+\ldots+{ }^{n} C_{n} \end{array} \] On multiplying…
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