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  • Help me please, - Ifthen K is equal to: - Binomial theorem and its simple applications - JEE Main

If \sum_{r=0}^{25}\left\{^{50}C_r\cdot\;^{50-r}C_{25-5} \right \} = K \left(^{50}C_{25} \right ) then K is equal to:

  • Option 1)

    (25)2

  • Option 2)

    224

  • Option 3)

    225

  • Option 4)

    225-1

Answers (1)

best_answer

 

Binomial Coefficients in Binomial Expansion -

In the expansion of \left ( x+a \right )^{n}, Binomial Coefficients are ^{n}c_{r}
 

- wherein

^{n}c_{r} is +ve .

\frac{n!}{r!\left ( n-r \right )!}

 

 

Sum of Binomial Coefficients -

(x+a)^{n}= ^{n}c_{0}x^{n}a^{0}+^{n}c_{r}x^{n-1}a+^{n}c_{2}x^{n-2}a^{2}+---

x= a= 1:

\therefore c_{0}+c_{1}+c_{2}+c_{3}+----= 2^{n}

-

\sum_{r=0}^{25}\left\{^{50}C_r\cdot\;^{50-r}C_{25-r} \right \}

from the concept learnt

=\sum_{r=0}^{25}\frac{50!}{r!(50-r)!}\times \frac{(50-r)!}{25!(25-r)!}

=\sum_{r=0}^{25}\frac{50!}{25!25!}\times \frac{25!}{25!(25-r)!}

=_{25}^{50}\textrm{C}\sum_{r=0}^{25}_{r}^{25}\textrm{C}=(2)^{25}\:\cdot \: _{25}^{50}\textrm{C}

Compare RHS and LHS 

K=2^{25}

 

 


Option 1)

(25)2

Option 2)

224

Option 3)

225

Option 4)

225-1

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