JEE Mains · Maths · STD 11 - 7. binomial theoram
Let \(0 \leq \mathrm{r} \leq \mathrm{n}\). If \({ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}+1}:{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}:{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}=55: 35: 21\), then \(2 n+5 r\) is equal to :
- A \(60\)
- B \(62\)
- C \(50\)
- D \(55\)
Answer & Solution
Correct Answer
(C) \(50\)
Step-by-step Solution
Detailed explanation
\( \frac{{ }^{n+1} C_r}{{ }^n C_r}=\frac{55}{35} \) \( \frac{(n+1) !}{(r+1) !(n-r)} ! \frac{r !(n-r) !}{n !}=\frac{11}{7} \) \( \frac{(n+1)}{r+1}=\frac{11}{7}\) \( 7 \mathrm{n}=4+11 \mathrm{r} \) \( \frac{{ }^n C_r}{{ }^{n-1} C_{r-1}}=\frac{35}{21} \)…
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