Let $n \in \mathbf{N}$ and $[x]$ denote the greatest integer less than or equal to $x$. If the sum of $(n+1)$ terms ${ }^{\mathrm{n}} \mathrm{C}_{0}, 3 \cdot{ }^{\mathrm{n}} \mathrm{C}_{1}, 5 \cdot{ }^{\mathrm{n}} \mathrm{C}_{2}, 7 \cdot{ }^{\mathrm{n}} \mathrm{C}_{3}, \ldots \ldots$ is equal to $2^{100} \cdot 101$
then $2\left[\frac{\mathrm{n}-1}{2}\right]$ is equal to_________.
$1 .{ }^{\mathrm{n}} \mathrm{C}_{0}+3 .{ }^{\mathrm{n}} \mathrm{C}_{1}+5 .{ }^{\mathrm{n}} \mathrm{C}_{2}+\ldots+(2 \mathrm{n}+1) .{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}$
$\mathrm{T}_{\mathrm{r}}=(2 \mathrm{r}+1)^{\mathrm{n}} \mathrm{C}$
$\mathrm{S}=\sum \mathrm{T}_{\mathrm{r}}$
$S=\Sigma(2 r+1)^{n} C_{r}=\Sigma 2 r^{n} C_{r}+\Sigma^{n} C_{r}$
$S=2\left(n \cdot 2^{n-1}\right)+2^{n}=2^{n}(n+1)$
$2^{n}(n+1)=2^{100} \cdot 101 \Rightarrow n=100$
$2\left[\frac{\mathrm{n}-1}{2}\right]=2\left[\frac{99}{2}\right]=98$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.