JEE Mains · Maths · STD 11 - 7. binomial theoram
Let \(m, n \in N\) and \(\operatorname{gcd}(2, n)=1\). If \(30\left(\begin{array}{l}30 \\ 0\end{array}\right)+29\left(\begin{array}{l}30 \\ 1\end{array}\right)+\ldots+2\left(\begin{array}{l}30 \\ 28\end{array}\right)+1\left(\begin{array}{l}30 \\ 29\end{array}\right)= n .2^{ m }\) then \(n + m\) is equal to (Here \(\left.\left(\begin{array}{l} n \\ k \end{array}\right)={ }^{ n } C _{ k }\right)\)
- A \(45\)
- B \(56\)
- C \(42\)
- D \(36\)
Answer & Solution
Correct Answer
(A) \(45\)
Step-by-step Solution
Detailed explanation
\(30\left({ }^{30} C _{0}\right)+29\left({ }^{30} C _{1}\right)+\ldots+2\left({ }^{30} C _{28}\right)+1\left({ }^{30} C _{29}\right)\) \(=30\left({ }^{30} C _{30}\right)+29\left({ }^{30} C _{29}\right)+\ldots \ldots+2\left({ }^{30} C _{2}\right)+1\left({ }^{30} C _{1}\right)\)…
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