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If $\sum_{r=0}^{20}{ }^{20+r} \mathrm{C}_r=\frac{p}{q}{ }^{40} \mathrm{C}_{20}$ and $\mathrm{GCD}$ of $(p, q)=1$, then $p^2-q^2=$
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1240
$\sum_{r=0}^{20}{ }^{20+r} \mathrm{C}_r=\frac{\mathrm{P}}{q}{ }^{40} \mathrm{C}_{20}$
$\sum_{r=0}^{20}{ }^{20+r} \mathrm{C}_r={ }^{20} \mathrm{C}_0+{ }^{21} \mathrm{C}_1+{ }^{22} \mathrm{C}_2+\ldots+{ }^{40} \mathrm{C}_{20}$
$\begin{aligned} & { }^{20} \mathrm{C}_0=1={ }^{21} \mathrm{C}_0 \\ & \text { replace }{ }^{20} \mathrm{C}_0 \text { with }{ }^{21} \mathrm{C}_0 \\ & { }^{21} \mathrm{C}_1+{ }^{21} \mathrm{C}_0={ }^{22} \mathrm{C}_1 \\ & { }^{22} \mathrm{C}_1+{ }^{22} \mathrm{C}_2={ }^{23} \mathrm{C}_2 \\ & { }^{23} \mathrm{C}_2+{ }^{23} \mathrm{C}_3={ }^{24} \mathrm{C}_3 \\ & \vdots \\ & { }^{39} \mathrm{C}_{19}+{ }^{39} \mathrm{C}_{20}={ }^{40} \mathrm{C}_{20} \\ & { }^{40} \mathrm{C}_{20}+{ }^{40} \mathrm{C}_{21}={ }^{41} \mathrm{C}_{21} \\ & =\frac{41}{21}{ }^{40} \mathrm{C}_{20} \\ & \therefore \mathrm{P}=41 \text { and } q=21 \\ & \mathrm{P}^2-q^2=41^2-21^2=1240\end{aligned}$
$\sum_{r=0}^{20}{ }^{20+r} \mathrm{C}_r={ }^{20} \mathrm{C}_0+{ }^{21} \mathrm{C}_1+{ }^{22} \mathrm{C}_2+\ldots+{ }^{40} \mathrm{C}_{20}$
$\begin{aligned} & { }^{20} \mathrm{C}_0=1={ }^{21} \mathrm{C}_0 \\ & \text { replace }{ }^{20} \mathrm{C}_0 \text { with }{ }^{21} \mathrm{C}_0 \\ & { }^{21} \mathrm{C}_1+{ }^{21} \mathrm{C}_0={ }^{22} \mathrm{C}_1 \\ & { }^{22} \mathrm{C}_1+{ }^{22} \mathrm{C}_2={ }^{23} \mathrm{C}_2 \\ & { }^{23} \mathrm{C}_2+{ }^{23} \mathrm{C}_3={ }^{24} \mathrm{C}_3 \\ & \vdots \\ & { }^{39} \mathrm{C}_{19}+{ }^{39} \mathrm{C}_{20}={ }^{40} \mathrm{C}_{20} \\ & { }^{40} \mathrm{C}_{20}+{ }^{40} \mathrm{C}_{21}={ }^{41} \mathrm{C}_{21} \\ & =\frac{41}{21}{ }^{40} \mathrm{C}_{20} \\ & \therefore \mathrm{P}=41 \text { and } q=21 \\ & \mathrm{P}^2-q^2=41^2-21^2=1240\end{aligned}$
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