Search any question & find its solution
Question:
Answered & Verified by Expert
If $\sum_{r=0}^{n} \frac{r+2}{r+1}^{n} C_{r}=\frac{2^{8}-1}{6},$ then $n=$
Options:
Solution:
2936 Upvotes
Verified Answer
The correct answer is:
5
$\sum_{r=0}^{n} \frac{r+2}{r+1}{ }^{n} C_{r}=\frac{2^{8}-1}{6}$
$$
\begin{aligned}
& \Rightarrow \sum_{r=0}^{n}\left[1+\frac{1}{r+1}\right]{ }^{n} C_{r}=\frac{2^{8}-1}{6} \\
& \Rightarrow 2^{n}+\sum_{r=0}^{n} \frac{1}{n+1} \cdot{ }^{n+1} C_{r+1}=\frac{2^{8}-1}{6} \\
\Rightarrow 2^{n}+& \frac{2^{n+1}-1}{n+1}=\frac{2^{8}-1}{6} \Rightarrow \frac{2^{n}(n+3)-1}{n+1}=\frac{2^{8}-1}{6} \\
\Rightarrow & \frac{2^{n}(n+1+2)-1}{n+1}=\frac{2^{5}(6+2)-1}{6}
\end{aligned}
$$
Comparing we get $\mathrm{n}+1=6 \Rightarrow \mathrm{n}=5$
$$
\begin{aligned}
& \Rightarrow \sum_{r=0}^{n}\left[1+\frac{1}{r+1}\right]{ }^{n} C_{r}=\frac{2^{8}-1}{6} \\
& \Rightarrow 2^{n}+\sum_{r=0}^{n} \frac{1}{n+1} \cdot{ }^{n+1} C_{r+1}=\frac{2^{8}-1}{6} \\
\Rightarrow 2^{n}+& \frac{2^{n+1}-1}{n+1}=\frac{2^{8}-1}{6} \Rightarrow \frac{2^{n}(n+3)-1}{n+1}=\frac{2^{8}-1}{6} \\
\Rightarrow & \frac{2^{n}(n+1+2)-1}{n+1}=\frac{2^{5}(6+2)-1}{6}
\end{aligned}
$$
Comparing we get $\mathrm{n}+1=6 \Rightarrow \mathrm{n}=5$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.