JEE Mains · Maths · STD 11 - 7. binomial theoram
If \(\frac{1}{n+1}{ }^n C_n+\frac{1}{n}{ }^n C_{n-1}+\ldots+\frac{1}{2}{ }^{ n } C _1+{ }^{ n } C _0=\frac{1023}{10}\) then \(n\) is equal to
- A \(6\)
- B \(9\)
- C \(8\)
- D \(7\)
Answer & Solution
Correct Answer
(B) \(9\)
Step-by-step Solution
Detailed explanation
\(\sum \limits_{ r =0}^{ n } \frac{{ }^{ n } C_{ r }}{ r +1}=\frac{1}{ n +1} \sum \limits_{ r =0}^{ n }{ }^{ n +1} C_{ r +1}\) \(=\frac{1}{ n +1}\left(2^{ n +1}-1\right)=\frac{1023}{10}\) \(n +1=10 \Rightarrow n =9\)
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