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QJEEM23Maa6f1: JEE Main 2023 Mathematics - Class 11 JEEM23Maa6f1 Solution

The value of 11!50!+13!48!+15!46!+….+149!2!+151!1!\frac{1}{1 ! 50 !}+\frac{1}{3 ! 48 !}+\frac{1}{5 ! 46 !}+\ldots .+\frac{1}{49 ! 2 !}+\frac{1}{51 ! 1 !} is :

Multiple Choice Options

  • A 25150!\frac{2^{51}}{50 !}
  • B 25151!\frac{2^{51}}{51 !}
  • C 25050!\frac{2^{50}}{50 !}
  • D 25051!\frac{2^{50}}{51 !}

Step-by-Step Verified Solution

amp;S=11!50!+13!48!+15!46!+……+149!2!+151!1!amp;=151!(51!1!50!+51!3!48!+51!5!46!+…..+51!49!2!+51!51!0!)amp;=151!(51C50+51C48+51C46+…….+51C2+51C0)amp;∵nC0+nC2+nC4+……=2n−1amp;∴S=25051!\begin{aligned}& \mathrm{S}=\frac{1}{1 ! 50 !}+\frac{1}{3 ! 48 !}+\frac{1}{5 ! 46 !}+\ldots \ldots+\frac{1}{49 ! 2 !}+\frac{1}{51 ! 1 !} \\\\& =\frac{1}{51 !}\left(\frac{51 !}{1 ! 50 !}+\frac{51 !}{3 ! 48 !}+\frac{51 !}{5 ! 46 !}+\ldots . .+\frac{51 !}{49 ! 2 !}+\frac{51 !}{51 ! 0 !}\right) \\\\& =\frac{1}{51 !}\left({ }^{51} C_{50}+{ }^{51} C_{48}+{ }^{51} C_{46}+\ldots \ldots .+{ }^{51} C_2+{ }^{51} C_0\right) \\\\& \because{ }^n C_0+{ }^n C_2+{ }^n C_4+\ldots \ldots=2^{n-1} \\\\& \therefore S=\frac{2^{50}}{51 !}\end{aligned}

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