Question:
If \({}^{20}C_1 + (2^2)\,{}^{20}C_2 + (3^2)\,{}^{20}C_3 + \ldots + (20^2)\,{}^{20}C_{20} = A(2^{\beta})\), then the ordered pair \((A, \beta)\) is equal to :
If \({}^{20}C_1 + (2^2)\,{}^{20}C_2 + (3^2)\,{}^{20}C_3 + \ldots + (20^2)\,{}^{20}C_{20} = A(2^{\beta})\), then the ordered pair \((A, \beta)\) is equal to :
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Write \(r^2 = r(r-1) + r\) and use \(r\,{}^nC_r = n\,{}^{n-1}C_{r-1}\) to reduce the sum to powers of 2.
Updated On: Nov 11, 2026
- \((420, 19)\)
- \((420, 18)\)
- \((380, 18)\)
- \((380, 19)\)
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The Correct Option is B
Solution and Explanation
The sum is \(S = \sum_{r=1}^{20} r^2\,{}^{20}C_r\). Write \(r^2 = r(r-1) + r\).
\(r(r-1)\,{}^{20}C_r = 20\cdot 19\,{}^{18}C_{r-2}\) and \(r\,{}^{20}C_r = 20\,{}^{19}C_{r-1}\).
So \(S = 380\sum {}^{18}C_{r-2} + 20\sum {}^{19}C_{r-1} = 380\cdot 2^{18} + 20\cdot 2^{19}\).
\(20\cdot 2^{19} = 40\cdot 2^{18}\), so \(S = (380+40)\,2^{18} = 420\cdot 2^{18}\).
Hence \(A = 420\), \(\beta = 18\).
Answer: (2) \((420, 18)\)
\(r(r-1)\,{}^{20}C_r = 20\cdot 19\,{}^{18}C_{r-2}\) and \(r\,{}^{20}C_r = 20\,{}^{19}C_{r-1}\).
So \(S = 380\sum {}^{18}C_{r-2} + 20\sum {}^{19}C_{r-1} = 380\cdot 2^{18} + 20\cdot 2^{19}\).
\(20\cdot 2^{19} = 40\cdot 2^{18}\), so \(S = (380+40)\,2^{18} = 420\cdot 2^{18}\).
Hence \(A = 420\), \(\beta = 18\).
Answer: (2) \((420, 18)\)
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