If 20C1+22 20C2+(3)2 20C3+….+(20)2 20C20=A2β then the ordered pair A,β=

∑r2⋅cr=∑r(r−1)⋅cr+∑r⋅cr=nn-12n-2+n2n-1=20⋅19⋅218+20⋅219=218(20)(19+2)=420×218

If 20C1+22 20C2+(3)2 20C3+….+(20)2 20C20=A2β then the ordered pair A,β=

∑r2⋅cr=∑r(r−1)⋅cr+∑r⋅cr=nn-12n-2+n2n-1=20⋅19⋅218+20⋅219=218(20)(19+2)=420×218

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If $^{20} C_{1} + (2^{2}) (^{20} C_{2}) + (3)^{2} (^{20} C_{3}) + \dots . + (20)^{2} (^{20} C_{20}) = A (2^{β})$ then the ordered pair $(A, β) =$

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$(420, 19)$

$(420, 18)$

$(380, 18)$

$(380, 19)$

answer is C.

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$\begin{array}{l} \sum r^{2} \cdot c_{r} = \sum r (r - 1) \cdot c_{r} + \sum r \cdot c_{r} = n (n - 1) 2^{n - 2} + n 2^{n - 1} \\ = 20 \cdot 19 \cdot 2^{18} + 20 \cdot 2^{19} = 2^{18} (20) (19 + 2) = 420 \times 2^{18} \end{array}$

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If 20C1+22 20C2+(3)2 20C3+….+(20)2 20C20=A2β then the ordered pair A,β=

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If $^{20} C_{1} + (2^{2}) (^{20} C_{2}) + (3)^{2} (^{20} C_{3}) + \dots . + (20)^{2} (^{20} C_{20}) = A (2^{β})$ then the ordered pair $(A, β) =$