7.Binomial Theorem
hard

$\sum_{ r =0}^{6}\left({ }^{6} C _{ r }{ }^{-6} C _{6- r }\right)$ ની કિમંત મેળવો.

A

$1124$

B

$1134$

C

$1024$

D

$924$

(JEE MAIN-2021)

Solution

$\sum_{ r =0}^{6}{ }^{6} C _{ r } \cdot{ }^{6} C _{6- r }$

$={ }^{6} C _{0} \cdot{ }^{6} C _{6}+{ }^{6} C _{1} \cdot{ }^{6} C _{5}+\ldots \ldots+{ }^{6} C _{6} \cdot{ }^{6} C _{0}$

Now,

$(1+x)^{6}(1+x)^{6}$

$=\left(\begin{array}{l}\left.{ }^{6} C_{0}+{ }^{6} C_{1} x+{ }^{6} C_{2} x^{2}+\ldots . . .+{ }^{6} C_{6} x^{6}\right) \\ \left({ }^{6} C_{0}+{ }^{6} C_{1} x+{ }^{6} C_{2} x^{2}+\ldots \ldots+{ }^{6} C_{6} x^{6}\right)\end{array}\right.$ 

Comparing coefficeint of $x^{6}$ both sides

$\begin{array}{l} { }^{6} C _{0} \cdot{ }^{6} C _{6}+{ }^{6} C _{1}+{ }^{6} C _{5}+\ldots \ldots .+{ }^{6} C _{6} \cdot{ }^{6} C _{0}={ }^{12} C _{6} \\ =924 \end{array}$

Standard 11
Mathematics

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