Skip to the question JEE Main › Mathematics › Binomial Theorem +4 marks−1 if incorrectNumerical pyq
Binomial Theorem: JEE Main Mathematics Question with Solution Suppose
∑ r = 0 2023 r 2 2023 C r = 2023 × α × 2 2022 \sum_{r=0}^{2023} r^2 \ ^{2023}\mathrm{C}_r = 2023 \times \alpha \times 2^{2022} ∑ r = 0 2023 r 2 2023 C r = 2023 × α × 2 2022 . Then the value of
α \alpha α is
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Choose one answer I’d rather see the answer directly Step-by-step solution View Correct answer
The value of α \alpha α Option analysis
Why each option works or fails
Step Working
01 given We are given the sum S = ∑ r = 0 n r 2 ( n r ) S = \sum_{r=0}^{n} r^2 \binom{n}{r} S = ∑ r = 0 n r 2 ( r n ) n = 2023 n = 2023 n = 2023 2023 × α × 2 2022 2023 \times \alpha \times 2^{2022} 2023 × α × 2 2022
02 goal Find the value of the integer constant α \alpha α
03 approach Split r 2 r^2 r 2 r ( r − 1 ) + r r(r - 1) + r r ( r − 1 ) + r r ( n r ) = n ( n − 1 r − 1 ) r \binom{n}{r} = n \binom{n-1}{r-1} r ( r n ) = n ( r − 1 n − 1 )
04 execute Expand r 2 = r ( r − 1 ) + r r^2 = r(r-1) + r r 2 = r ( r − 1 ) + r ∑ r = 0 n r 2 ( n r ) = ∑ r = 2 n r ( r − 1 ) ( n r ) + ∑ r = 1 n r ( n 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} ∑ r = 0 n r 2 ( r n ) = ∑ r = 2 n r ( r − 1 ) ( r n ) + ∑ r = 1 n r ( r n ) r ( r − 1 ) ( n r ) = n ( n − 1 ) ( n − 2 r − 2 ) r(r-1)\binom{n}{r} = n(n-1)\binom{n-2}{r-2} r ( r − 1 ) ( r n ) = n ( n − 1 ) ( r − 2 n − 2 ) r ( n r ) = n ( n − 1 r − 1 ) r\binom{n}{r} = n\binom{n-1}{r-1} r ( r n ) = n ( r − 1 n − 1 ) S = n ( n − 1 ) ∑ r = 2 n ( n − 2 r − 2 ) + n ∑ r = 1 n ( n − 1 r − 1 ) = n ( n − 1 ) 2 n − 2 + n 2 n − 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} S = n ( n − 1 ) ∑ r = 2 n ( r − 2 n − 2 ) + n ∑ r = 1 n ( r − 1 n − 1 ) = n ( n − 1 ) 2 n − 2 + n 2 n − 1 n = 2023 n = 2023 n = 2023 S = 2023 × 2022 × 2 2021 + 2023 × 2 2022 = 2023 × 2 2021 ( 2022 + 2 ) = 2023 × 2 2021 × 2024 S = 2023 \times 2022 \times 2^{2021} + 2023 \times 2^{2022} = 2023 \times 2^{2021} (2022 + 2) = 2023 \times 2^{2021} \times 2024 S = 2023 × 2022 × 2 2021 + 2023 × 2 2022 = 2023 × 2 2021 ( 2022 + 2 ) = 2023 × 2 2021 × 2024 2023 × α × 2 2022 2023 \times \alpha \times 2^{2022} 2023 × α × 2 2022 S = 2023 × 2024 2 × 2 2022 = 2023 × 1012 × 2 2022 S = 2023 \times \frac{2024}{2} \times 2^{2022} = 2023 \times 1012 \times 2^{2022} S = 2023 × 2 2024 × 2 2022 = 2023 × 1012 × 2 2022 α = 1012 \alpha = 1012 α = 1012
✓ verify Verify via general formula: ∑ r = 0 n r 2 ( n r ) = n ( n + 1 ) 2 n − 2 \sum_{r=0}^n r^2 \binom{n}{r} = n(n+1)2^{n-2} ∑ r = 0 n r 2 ( r n ) = n ( n + 1 ) 2 n − 2 n ( n + 1 ) 2 n − 2 = 2023 × 2024 × 2 2021 = 2023 × 1012 × 2 2022 n(n+1)2^{n-2} = 2023 \times 2024 \times 2^{2021} = 2023 \times 1012 \times 2^{2022} n ( n + 1 ) 2 n − 2 = 2023 × 2024 × 2 2021 = 2023 × 1012 × 2 2022 α = 1012 \alpha = 1012 α = 1012
Your next move We think you should solve this next ✓ Source and academic review↓
Question type Numerical
Exam relevance JEE Main · Mathematics
Concepts assessed Mathematics
Academic status Reviewed by official_key
Source pyq
Editorial review 9 September 2026 Quick checks
Students also ask Why do we split r 2 r^2 r 2 r ( r − 1 ) + r r(r-1) + r r ( r − 1 ) + r r r r Because the binomial coefficient absorption identity works on consecutive decreasing factors r ( r − 1 ) … r(r-1)\dots r ( r − 1 ) … ( n r ) \binom{n}{r} ( r n )
Can we directly remember the result ∑ r 2 ( n r ) = n ( n + 1 ) 2 n − 2 \sum r^2 \binom{n}{r} = n(n+1)2^{n-2} ∑ r 2 ( r n ) = 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 \alpha α
Why each option works or fails Step-by-step solution given: We are given the sum S = ∑ r = 0 n r 2 ( n r ) S = \sum_{r=0}^{n} r^2 \binom{n}{r} S = ∑ r = 0 n r 2 ( r n ) n = 2023 n = 2023 n = 2023 2023 × α × 2 2022 2023 \times \alpha \times 2^{2022} 2023 × α × 2 2022 goal: Find the value of the integer constant α \alpha α approach: Split r 2 r^2 r 2 r ( r − 1 ) + r r(r - 1) + r r ( r − 1 ) + r r ( n r ) = n ( n − 1 r − 1 ) r \binom{n}{r} = n \binom{n-1}{r-1} r ( r n ) = n ( r − 1 n − 1 ) execute: Expand r 2 = r ( r − 1 ) + r r^2 = r(r-1) + r r 2 = r ( r − 1 ) + r ∑ r = 0 n r 2 ( n r ) = ∑ r = 2 n r ( r − 1 ) ( n r ) + ∑ r = 1 n r ( n 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} ∑ r = 0 n r 2 ( r n ) = ∑ r = 2 n r ( r − 1 ) ( r n ) + ∑ r = 1 n r ( r n ) r ( r − 1 ) ( n r ) = n ( n − 1 ) ( n − 2 r − 2 ) r(r-1)\binom{n}{r} = n(n-1)\binom{n-2}{r-2} r ( r − 1 ) ( r n ) = n ( n − 1 ) ( r − 2 n − 2 ) r ( n r ) = n ( n − 1 r − 1 ) r\binom{n}{r} = n\binom{n-1}{r-1} r ( r n ) = n ( r − 1 n − 1 ) S = n ( n − 1 ) ∑ r = 2 n ( n − 2 r − 2 ) + n ∑ r = 1 n ( n − 1 r − 1 ) = n ( n − 1 ) 2 n − 2 + n 2 n − 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} S = n ( n − 1 ) ∑ r = 2 n ( r − 2 n − 2 ) + n ∑ r = 1 n ( r − 1 n − 1 ) = n ( n − 1 ) 2 n − 2 + n 2 n − 1 n = 2023 n = 2023 n = 2023 S = 2023 × 2022 × 2 2021 + 2023 × 2 2022 = 2023 × 2 2021 ( 2022 + 2 ) = 2023 × 2 2021 × 2024 S = 2023 \times 2022 \times 2^{2021} + 2023 \times 2^{2022} = 2023 \times 2^{2021} (2022 + 2) = 2023 \times 2^{2021} \times 2024 S = 2023 × 2022 × 2 2021 + 2023 × 2 2022 = 2023 × 2 2021 ( 2022 + 2 ) = 2023 × 2 2021 × 2024 2023 × α × 2 2022 2023 \times \alpha \times 2^{2022} 2023 × α × 2 2022 S = 2023 × 2024 2 × 2 2022 = 2023 × 1012 × 2 2022 S = 2023 \times \frac{2024}{2} \times 2^{2022} = 2023 \times 1012 \times 2^{2022} S = 2023 × 2 2024 × 2 2022 = 2023 × 1012 × 2 2022 α = 1012 \alpha = 1012 α = 1012 verify: Verify via general formula: ∑ r = 0 n r 2 ( n r ) = n ( n + 1 ) 2 n − 2 \sum_{r=0}^n r^2 \binom{n}{r} = n(n+1)2^{n-2} ∑ r = 0 n r 2 ( r n ) = n ( n + 1 ) 2 n − 2 n ( n + 1 ) 2 n − 2 = 2023 × 2024 × 2 2021 = 2023 × 1012 × 2 2022 n(n+1)2^{n-2} = 2023 \times 2024 \times 2^{2021} = 2023 \times 1012 \times 2^{2022} n ( n + 1 ) 2 n − 2 = 2023 × 2024 × 2 2021 = 2023 × 1012 × 2 2022 α = 1012 \alpha = 1012 α = 1012 Shortcut: When to use it: Fastest when comfortable differentiating ( 1 + x ) n (1+x)^n ( 1 + x ) n
given: Consider the expansion ( 1 + x ) n = ∑ r = 0 n ( n r ) x r (1+x)^n = \sum_{r=0}^n \binom{n}{r} x^r ( 1 + x ) n = ∑ r = 0 n ( r n ) x r n = 2023 n = 2023 n = 2023
goal: Obtain ∑ r = 0 n r 2 ( n r ) \sum_{r=0}^n r^2 \binom{n}{r} ∑ r = 0 n r 2 ( r n ) x d d x x \frac{d}{dx} x d x d
approach: Differentiate once, multiply by x x x x = 1 x = 1 x = 1
execute: First derivative:
n ( 1 + x ) n − 1 = ∑ r = 1 n r ( n r ) x r − 1 n(1+x)^{n-1} = \sum_{r=1}^n r \binom{n}{r} x^{r-1} n ( 1 + x ) n − 1 = ∑ r = 1 n r ( r n ) x r − 1 x x x n x ( 1 + x ) n − 1 = ∑ r = 1 n r ( n r ) x r n x (1+x)^{n-1} = \sum_{r=1}^n r \binom{n}{r} x^r n x ( 1 + x ) n − 1 = ∑ r = 1 n r ( r n ) x r x x x n ( 1 + x ) n − 1 + n ( n − 1 ) x ( 1 + x ) n − 2 = ∑ r = 1 n r 2 ( n r ) x r − 1 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} n ( 1 + x ) n − 1 + n ( n − 1 ) x ( 1 + x ) n − 2 = ∑ r = 1 n r 2 ( r n ) x r − 1 x = 1 x = 1 x = 1 ∑ r = 0 n r 2 ( 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 \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} ∑ r = 0 n r 2 ( r n ) = n 2 n − 1 + n ( n − 1 ) 2 n − 2 = n 2 n − 2 [ 2 + n − 1 ] = n ( n + 1 ) 2 n − 2 n = 2023 n = 2023 n = 2023 S = 2023 × 2024 × 2 2021 = 2023 × 1012 × 2 2022 S = 2023 \times 2024 \times 2^{2021} = 2023 \times 1012 \times 2^{2022} S = 2023 × 2024 × 2 2021 = 2023 × 1012 × 2 2022 2023 × α × 2 2022 2023 \times \alpha \times 2^{2022} 2023 × α × 2 2022 α = 1012 \alpha = 1012 α = 1012
verify: For n = 1 n=1 n = 1 ∑ r = 0 1 r 2 ( 1 r ) = 0 + 1 = 1 \sum_{r=0}^1 r^2 \binom{1}{r} = 0 + 1 = 1 ∑ r = 0 1 r 2 ( r 1 ) = 0 + 1 = 1 1 ( 2 ) 2 − 1 = 1 1(2)2^{-1} = 1 1 ( 2 ) 2 − 1 = 1