JEE MainMathematics

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Evaluating Binomial Coefficient Sums with Quadratic Weights

Suppose sum from r=0 to 2023 of r^2 ^2023C_r = 2023 × α × 2^2022. Then the value of α is

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Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

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Students also ask

Why do we split

r^{2}

into

r (r - 1) + r

instead of just cancelling

r

Because the binomial coefficient absorption identity works on consecutive decreasing factors $r (r - 1) \dots$ , which match the falling factorials in $(\binom{n}{r})$ .

Can we directly remember the result

\sum r^{2} (\binom{n}{r}) = n (n + 1) 2^{n - 2}

Yes, it is a standard JEE result that can be directly quoted and used in numerical questions.

Answer

The value of $α$ is 1012.

Why each option works or fails

Step-by-step solution

Given: We are given the sum $S = \sum_{r = 0}^{n} r^{2} (\binom{n}{r})$ where $n = 2023$ , and it is equated to $2023 \times α \times 2^{2022}$ .
Goal: Find the value of the integer constant $α$ .
Approach: Split $r^{2}$ into $r (r - 1) + r$ to apply the binomial absorption identity $r (\binom{n}{r}) = n (\binom{n - 1}{r - 1})$ iteratively.
Execute: Expand $r^{2} = r (r - 1) + r$ : $\sum_{r = 0}^{n} r^{2} (\binom{n}{r}) = \sum_{r = 2}^{n} r (r - 1) (\binom{n}{r}) + \sum_{r = 1}^{n} r (\binom{n}{r})$ Using $r (r - 1) (\binom{n}{r}) = n (n - 1) (\binom{n - 2}{r - 2})$ and $r (\binom{n}{r}) = n (\binom{n - 1}{r - 1})$ : $S = n (n - 1) \sum_{r = 2}^{n} (\binom{n - 2}{r - 2}) + n \sum_{r = 1}^{n} (\binom{n - 1}{r - 1}) = n (n - 1) 2^{n - 2} + n 2^{n - 1}$ Substitute $n = 2023$ : $S = 2023 \times 2022 \times 2^{2021} + 2023 \times 2^{2022} = 2023 \times 2^{2021} (2022 + 2) = 2023 \times 2^{2021} \times 2024$ Rewrite in the form $2023 \times α \times 2^{2022}$ : $S = 2023 \times \frac{2024}{2} \times 2^{2022} = 2023 \times 1012 \times 2^{2022}$ Hence, $α = 1012$ .
Verify: Verify via general formula: $\sum_{r = 0}^{n} r^{2} (\binom{n}{r}) = n (n + 1) 2^{n - 2}$ . Here, $n (n + 1) 2^{n - 2} = 2023 \times 2024 \times 2^{2021} = 2023 \times 1012 \times 2^{2022}$ . Matching coefficients confirms $α = 1012$ .

Shortcut:

When to use it: Fastest when comfortable differentiating $(1 + x)^{n}$ twice

Given: Consider the expansion $(1 + x)^{n} = \sum_{r = 0}^{n} (\binom{n}{r}) x^{r}$ with $n = 2023$ .

Goal: Obtain $\sum_{r = 0}^{n} r^{2} (\binom{n}{r})$ by applying the operator $x \frac{d}{d x}$ .

Approach: Differentiate once, multiply by $x$ , then differentiate again and set $x = 1$ .

Execute: First derivative: $n (1 + x)^{n - 1} = \sum_{r = 1}^{n} r (\binom{n}{r}) x^{r - 1}$ Multiply by $x$ : $n x (1 + x)^{n - 1} = \sum_{r = 1}^{n} r (\binom{n}{r}) x^{r}$ Differentiate with respect to $x$ : $n (1 + x)^{n - 1} + n (n - 1) x (1 + x)^{n - 2} = \sum_{r = 1}^{n} r^{2} (\binom{n}{r}) x^{r - 1}$ Put $x = 1$ : $\sum_{r = 0}^{n} r^{2} (\binom{n}{r}) = n 2^{n - 1} + n (n - 1) 2^{n - 2} = n 2^{n - 2} [2 + n - 1] = n (n + 1) 2^{n - 2}$ For $n = 2023$ : $S = 2023 \times 2024 \times 2^{2021} = 2023 \times 1012 \times 2^{2022}$ Comparing with $2023 \times α \times 2^{2022}$ , we obtain $α = 1012$ .

Verify: For $n = 1$ : $\sum_{r = 0}^{1} r^{2} (\binom{1}{r}) = 0 + 1 = 1$ . The formula gives $1 (2) 2^{- 1} = 1$ . It holds universally.

Evaluating Binomial Coefficient Sums with Quadratic Weights

What feels right?

The value of $α$ is 1012.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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