MHT CET · Maths · Binomial Theorem
If \({ }^{\mathrm{n}} \mathrm{C}_0+\frac{1}{2}{ }^{\mathrm{n}} \mathrm{C}_1+\frac{1}{3}{ }^{\mathrm{n}} \mathrm{C}_2+\ldots \ldots \ldots \ldots . . \cdot \frac{1}{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-1}+\frac{1}{\mathrm{n}+1}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}=\frac{1023}{10}\) then \(\mathrm{n}=\)
- A 7
- B 8
- C 9
- D 10
Answer & Solution
Correct Answer
(C) 9
Step-by-step Solution
Detailed explanation
\( \sum_{k=0}^{n} \frac{1}{k+1} { }^{\mathrm{n}} \mathrm{C}_k = \frac{2^{n+1}-1}{n+1} \) \( \frac{2^{n+1}-1}{n+1} = \frac{1023}{10} \)
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