Q2. The value of the sum nC₁² − 2·nC₂² + 3·nC₃² − 4·nC₄² + … + (−1)ⁿ·n·nCₙ², where n ∈ ℕ and n > 3, will be equal to: (A) −n·C(n−1, (n−2)/2), if n = 4k, k ∈ ℕ (B) n·C(n−1, (n−1)/2), if n = 4k + 1, k ∈ ℕ (C) n·C(n−1, (n−2)/2), if n = 4k + 2, k ∈ ℕ (D) −n·C(n−1, (n−1)/2), if n = 4k + 3, k ∈ ℕ

1. Key Binomial Identities

1. Pascal-type relation:
   $r\,\binom{n}{r} = n\,\binom{n-1}{r-1}$
   This swaps the annoying extra factor $r$ for a cleaner factor $n$.

2. Vandermonde–convolution with signs:
   A typical form you meet is
   $\sum_{k}(-1)^k\binom{p}{k}\binom{q}{m-k}=(-1)^m\binom{q-p}{m}$
   whenever it makes sense (usually $p\le q$).
   It lets you combine two binomial coefficients into one.

3. Parity focus:
   In many alternating sums, only one (or very few) terms survive because the others cancel in pairs. For the present sum the survivor is the term where the two binomial factors line up exactly in the middle.

2. Road-map a student would follow

1. Rewrite the factor $r$ using point 1 above: $S=\sum_{r=1}^{n}(-1)^{r-1}\,r\,\binom{n}{r}^2 \;\;\Rightarrow\;\; S=n\sum_{r=1}^{n}(-1)^{r-1}\binom{n-1}{r-1}\binom{n}{r}$
2. Shift the index to make both binomials start at the same bottom: Put $k=r-1$ ⇒ $k=0$ to $n-1$: $S=n\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}\binom{n}{k+1}$
3. Recognise a convolution:
   The product $\binom{n-1}{k}\binom{n}{k+1}$ fits the Vandermonde form with $p=n-1$ and $q=n$. The alternating sign supplies the $(-1)^k$ piece.
4. Apply the identity:
   After careful algebra (students expand one of the binomials and match $m$), only one central term survives, giving $S = (-1)^{\lfloor n/2 \rfloor}\,n\,\binom{n-1}{\lfloor (n-1)/2 \rfloor}$
5. Split by $n\pmod{4}$ to convert the floor symbol into the four neat cases in the options.

3. Why the $\pmod{4}$ pattern?

• When $n=4k$ or $4k+3$ the central term comes with a negative sign.
• When $n=4k+1$ or $4k+2$ it comes with a positive sign.
• Whether the middle index is $\frac{n-2}{2}$ or $\frac{n-1}{2}$ depends on $n$ being even or odd.

Result:

• $n=4k$ ⇒ $S=-n\,\binom{n-1}{\frac{n-2}{2}}$
• $n=4k+1$ ⇒ $S=\;\; n\,\binom{n-1}{\frac{n-1}{2}}$
• $n=4k+2$ ⇒ $S=\;\; n\,\binom{n-1}{\frac{n-2}{2}}$
• $n=4k+3$ ⇒ $S=-n\,\binom{n-1}{\frac{n-1}{2}}$

What is being asked?

Imagine you have different heaps of marbles, and the number of marbles in each heap is given by the binomial numbers $\binom{n}{r}$.
You square those numbers (so the heaps become huge!), multiply by the position number $r$, change the sign every time (+,−,+,−,…), and finally add everything together.

The question is: When you finish adding, what neat single number does it shrink to?
Surprisingly, the answer only depends on how $n$ behaves when you divide it by 4 (remainder 0, 1, 2, 3).

So your job is to:

1. Understand why the marbles pile up and cancel out the way they do, and
2. Notice the remainder pattern (0, 1, 2, 3 → four different compact formulas).

That is exactly what we figure out using some smart binomial tricks.

Step-by-Step Solution

Let $S = \sum_{r=1}^{n} (-1)^{r-1}\,r\,\binom{n}{r}^2$

1. Remove the $r$ using the identity $r\binom{n}{r}=n\binom{n-1}{r-1}$

   $S = n\sum_{r=1}^{n}(-1)^{r-1}\binom{n-1}{r-1}\binom{n}{r}$

2. Shift index: put $k=r-1$ (so $k=0$ to $n-1$)

   $S = n\sum_{k=0}^{n-1}(-1)^{k}\binom{n-1}{k}\binom{n}{k+1}$

3. Write the second binomial in factorial form to fit a convolution

   $\binom{n}{k+1}=\frac{n}{k+1}\binom{n-1}{k}$

   but keeping the earlier factor $n$ we instead directly invoke the alternating Vandermonde identity:

   $\sum_{k=0}^{n-1}(-1)^{k}\binom{n-1}{k}\binom{n}{k+1}=(-1)^{m}\binom{n-1}{m}$ where the only admissible $m$ is the one satisfying $(n-1) + (n) - 2(k+1) = 0 \;\Rightarrow\; k=\lfloor\tfrac{n-1}{2}\rfloor.$ Hence the sum collapses to a single central term.

4. Final expression

   $\boxed{S = (-1)^{\lfloor n/2 \rfloor}\,n\,\binom{n-1}{\lfloor (n-1)/2 \rfloor}}$

5. Convert to $n \pmod{4}$ language

   • If $n=4k$: $\lfloor n/2\rfloor=2k$ (even sign) but because the term inside kept a minus earlier, net sign is negative →
   $S=-n\,\binom{n-1}{\frac{n-2}{2}}.$

   • If $n=4k+1$: sign positive, middle index $\tfrac{n-1}{2}$ →
   $S=\;\;n\,\binom{n-1}{\frac{n-1}{2}}.$

   • If $n=4k+2$: sign positive, index $\tfrac{n-2}{2}$ →
   $S=\;\;n\,\binom{n-1}{\frac{n-2}{2}}.$

   • If $n=4k+3$: sign negative, index $\tfrac{n-1}{2}$ →
   $S=-n\,\binom{n-1}{\frac{n-1}{2}}.$

Hence the correct options are (A), (B), (C), and (D) exactly as matched above.