TS EAMCET · Maths · Binomial Theorem
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 \(p^2-q^2=\)
- A 1302
- B 1220
- C 1240
- D 1364
Answer & Solution
Correct Answer
(C) 1240
Step-by-step Solution
Detailed explanation
\(\sum_{r=0}^{20}{ }^{20+r} \mathrm{C}_r=\frac{\mathrm{P}}{q}{ }^{40} \mathrm{C}_{20}\) \(\sum_{r=0}^{20}{ }^{20+r} \mathrm{C}_r={ }^{20} \mathrm{C}_0+{ }^{21} \mathrm{C}_1+{ }^{22} \mathrm{C}_2+\ldots+{ }^{40} \mathrm{C}_{20}\)…
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