Let n ∈ N and [x] denote the greatest integer less than or equal to...

Let $S = C0 n + 3 C1 n + 5 . C2 n + \ldots + \left(2n + 1\right) \cdot Cn n$

Now, ${T}_{r} = {\left(2r + 1\right)}^{n}{C}_{r}$

Therefore, $S = Σ{T}_{r}$

$S = Σ{\left(2r + 1\right)}^{n}{C}_{r}$

$= Σ2r\left(n{C}_{r}\right) + Σ\left(n{C}_{r}\right)$

$⇒ S = 2\left(n \cdot {2}^{n - 1}\right) + {2}^{n} = {2}^{n}\left(n + 1\right)$

$⇒ {2}^{n}\left(n + 1\right) = {2}^{100} . 101$

$⇒ n = 100$

Now, $2\left[\dfrac{n - 1}{2}\right] = 2\left[\dfrac{99}{2}\right] = 98 .$