Let $$n \in \mathbf{N}$$ and $$[x]$$ denote the greatest integer less than or equal to $$x$$. If the sum of $$(n + 1)$$ terms of $$^nC_0, 3 \cdot ^nC_1, 5 \cdot ^nC_2, 7 \cdot ^nC_3, \ldots$$ is equal to $$2^{100} \cdot 101$$, then $$2\left[\frac{n-1}{2}\right]$$ is equal to

We begin with the given series

$$^nC_0,\; 3\cdot {}^nC_1,\; 5\cdot {}^nC_2,\; 7\cdot {}^nC_3,\; \ldots,\;(n+1)\text{ terms}$$

Every coefficient in front of the binomial term is an odd number. In fact the $$k^{\text{th}}$$ term (starting with $$k=0$$) is

$$ (2k+1)\;{}^nC_k. $$

Therefore the required sum can be written compactly as

$$ S \;=\; \sum_{k=0}^{n} (2k+1)\;{}^nC_k. $$

We are told that this sum equals $$2^{100}\cdot 101$$, i.e.

$$ \sum_{k=0}^{n} (2k+1)\;{}^nC_k \;=\; 2^{100}\cdot 101. \quad -(1) $$

Now let us evaluate the left‐hand side purely algebraically. We separate the part involving $$k$$ and the part not involving $$k$$:

$$ \sum_{k=0}^{n} (2k+1)\;{}^nC_k \;=\; 2\sum_{k=0}^{n} k\;{}^nC_k \;+\; \sum_{k=0}^{n} {}^nC_k. $$

For the two standard binomial sums that appear here, we first state the well-known identities:

• Binomial sum $$\displaystyle\sum_{k=0}^{n} {}^nC_k = 2^n.$$
• Weighted binomial sum $$\displaystyle\sum_{k=0}^{n} k\;{}^nC_k = n\,2^{\,n-1}.$$

Using these, we obtain

$$ \begin{aligned} \sum_{k=0}^{n} (2k+1)\;{}^nC_k &= 2\Bigl(n\,2^{\,n-1}\Bigr) \;+\; 2^n \\ &= n\,2^{\,n} \;+\; 2^{\,n} \\ &= (n+1)\,2^{\,n}. \\ \end{aligned} $$

Hence the equation (1) becomes

$$ (n+1)\,2^{\,n} \;=\; 2^{100}\cdot 101. \quad -(2) $$

To compare the two sides, we isolate the odd and even (power-of-two) parts. Every positive integer can be written uniquely as $$\text{(odd part)}\times 2^{\text{(power of 2 present)}}.$$ Let us denote by $$v_2(m)$$ the exponent of the highest power of $$2$$ dividing $$m$$.

Write $$n+1 = 2^{\,r}\,m,$$ where $$m$$ is odd and $$r = v_2(n+1).$$ Then the left-hand side of (2) has a total power of two equal to $$n + r,$$ and its odd part equals $$m.$$ Comparing with the right-hand side, whose power of two is $$100$$ and whose odd part is $$101,$$ we obtain two simultaneous requirements:

$$ \begin{cases} m \;=\; 101,\\[4pt] n + r \;=\; 100. \end{cases} $$

But since $$m=101,$$ we have

$$ n+1 = 101\;2^{\,r}. \quad -(3) $$

Substituting $$n = 101\;2^{\,r} - 1$$ into the second requirement $$n + r = 100,$$ we get

$$ 101\;2^{\,r} - 1 + r \;=\; 100 \;\;\Longrightarrow\;\; 101\;2^{\,r} + r \;=\; 101. \quad -(4) $$

Because $$2^{\,r} \ge 1$$ and $$r \ge 0,$$ the left-hand side of (4) increases rapidly with $$r.$$ Testing non-negative integers:

• If $$r = 0$$ then $$101\cdot 1 + 0 = 101,$$ which satisfies (4).
• If $$r \ge 1$$ then $$101\;2^{\,r} \ge 202,$$ so the left-hand side exceeds $$101,$$ violating (4).

Thus $$r = 0$$ is the only admissible value. Feeding this back into (3) yields

$$ n+1 = 101\cdot 2^{\,0} = 101 \;\;\Longrightarrow\;\; n = 100. $$

The quantity asked for in the problem is

$$ 2\left[\frac{n-1}{2}\right]. $$

With $$n = 100,$$ we have

$$ \frac{n-1}{2} = \frac{100 - 1}{2} = \frac{99}{2} = 49.5. $$

The greatest integer function gives $$\left[\frac{99}{2}\right] = 49.$$

Multiplying by $$2$$ we obtain

$$ 2 \times 49 = 98. $$

So, the answer is $$98$$.