🛠️ JEE➗ Maths

The value of \(\frac{{}^{100}C_{50}}{51}+\frac{{}^{100}C_{51}}{52}+\ldots +\frac{{}^{100}C_{100}}{101}\) is:

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The value of \(\frac{{}^{100}C_{50}}{51}+\frac{{}^{100}C_{51}}{52}+\ldots +\frac{{}^{100}C_{100}}{101}\) is:

a

\(\frac{{2}^{100}}{100}\)

b

\(\frac{{2}^{101}}{101}\)

c

\(\frac{{2}^{101}}{100}\)

d

\(\frac{{2}^{100}}{101}\)

✓ Correct answer: d)

\(\frac{{2}^{100}}{101}\)

Explanation

\(S=\sum _{r=50}^{100}\frac{{}^{100}C_{r}}{r+1}=\sum _{r=50}^{100}\frac{1}{r+1}\cdot \frac{r+1}{101}\cdot {}^{101}C_{r+1}\\ S=\frac{1}{101}\sum _{r=50}^{100}{}^{101}C_{r+1}\\ ∵\sum _{r=0}^{n}{}^{n}C_{r}={2}^{n},{}^{n}C_{r}={}^{n}C_{n-r}\\ \Rightarrow \sum _{r=0}^{50}{}^{101}C_{r}=\sum _{r=51}^{101}{}^{101}C_{r}=\frac{{2}^{101}}{2}={2}^{100}\\ ∴S=\frac{1}{101}\times \frac{{2}^{101}}{2}=\frac{{2}^{100}}{101}\)

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