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JEE Main 202324 JanuaryMorning Shift
Exam
JEE Main
Year
2023
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2023 · 24 January · Morning Shift · Mathematics · Binomial Theorem

Suppose ∑ _ r = 0 ^ 2023 r ^ 2 · Cr 2023 = 2023 × α × 2 ^ 2022 ,...

Suppose ∑r=02023r2⋅Cr2023=2023×α×22022\mathop{∑}_{r = 0}^{2023}{r}^{2} \cdot Cr 2023 = 2023 \times \alpha \times {2}^{2022}, then the value of α\alpha is

Correct answer

0

Step-by-step solution

Given,

∑r=02023r2⋅Cr2023=2023×α×22022\mathop{∑}_{r = 0}^{2023}{r}^{2} \cdot Cr 2023 = 2023 \times \alpha \times {2}^{2022}

Now taking L.H.S we get,

∑r=02023r2⋅Cr2023=∑r=12023r2⋅2023r⋅Cr−12022\mathop{∑}_{r = 0}^{2023}{r}^{2} \cdot Cr 2023 = \mathop{∑}_{r = 1}^{2023}{r}^{2} \cdot \dfrac{2023}{r} \cdot Cr - 1 2022

⇒∑r=02023r2⋅Cr2023=∑r=12023r⋅2023⋅Cr−12022⇒ \mathop{∑}_{r = 0}^{2023}{r}^{2} \cdot Cr 2023 = \mathop{∑}_{r = 1}^{2023}r \cdot 2023 \cdot Cr - 1 2022

⇒∑r=02023r2⋅Cr2023=2023∑r=12023(r−1+1)⋅Cr−12022⇒ \mathop{∑}_{r = 0}^{2023}{r}^{2} \cdot Cr 2023 = 2023\mathop{∑}_{r = 1}^{2023}\left(r - 1 + 1\right) \cdot Cr - 1 2022

⇒∑r=02023r2⋅Cr2023=2023[∑r=12023(r−1)⋅Cr−12022+∑r=12023Cr−12022]⇒ \mathop{∑}_{r = 0}^{2023}{r}^{2} \cdot Cr 2023 = 2023\left[\mathop{∑}_{r = 1}^{2023}\left(r - 1\right) \cdot Cr - 1 2022 + \mathop{∑}_{r = 1}^{2023} Cr - 1 2022\right]

⇒∑r=02023r2⋅Cr2023=2023[2022∑r=22023Cr−22021+∑r=12023Cr−12022]⇒ \mathop{∑}_{r = 0}^{2023}{r}^{2} \cdot Cr 2023 = 2023\left[2022\mathop{∑}_{r = 2}^{2023} Cr - 2 2021 + \mathop{∑}_{r = 1}^{2023} Cr - 1 2022\right]

⇒∑r=02023r2⋅Cr2023=2023[2022×22021+22022]⇒ \mathop{∑}_{r = 0}^{2023}{r}^{2} \cdot Cr 2023 = 2023\left[2022 \times {2}^{2021} + {2}^{2022}\right]

⇒∑r=02023r2⋅Cr2023=2023×22022[1011+1]⇒ \mathop{∑}_{r = 0}^{2023}{r}^{2} \cdot Cr 2023 = 2023 \times {2}^{2022}\left[1011 + 1\right]

⇒∑r=02023r2⋅Cr2023=2023×1012×22022⇒ \mathop{∑}_{r = 0}^{2023}{r}^{2} \cdot Cr 2023 = 2023 \times 1012 \times {2}^{2022}

Now on comparing with ∑r=02023r2⋅Cr2023=2023×α×22022\mathop{∑}_{r = 0}^{2023}{r}^{2} \cdot Cr 2023 = 2023 \times \alpha \times {2}^{2022} we get,

α=1012\alpha = 1012

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