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Q. If the value of the sum $29\left(\left(\_{}^{30}C\right)_{0}\right)+28\left(\left(\_{}^{30}C\right)_{1}\right)+27\left(\left(\_{}^{30}C\right)_{2}\right)+.\ldots \ldots .+1\left(\left(\_{}^{30}C\right)_{28}\right)+0\cdot \left(\left(\_{}^{30}C\right)_{29}\right)-\left(\left(\_{}^{30}C\right)_{30}\right)$ is equal to $K\cdot 2^{32},$ then the value of $K$ is equal to

NTA AbhyasNTA Abhyas 2020Binomial Theorem

Solution:

$S=29\left(\left(\_{}^{30}C\right)_{0}\right)+28\left(\left(\_{}^{30}C\right)_{1}\right)+27\left(\left(\_{}^{30}C\right)_{2}\right)+.\ldots \ldots .+1\left(\left(\_{}^{30}C\right)_{28}\right)+0\cdot \left(\left(\_{}^{30}C\right)_{29}\right)-\left(\left(\_{}^{30}C\right)_{30}\right)$
Also,
$S=-\left(\left(\_{}^{30}C\right)_{0}\right)+0\cdot \left(\left(\_{}^{30}C\right)_{1}\right)+1\left(\left(\_{}^{30}C\right)_{2}\right)+.\ldots \ldots .+27\left(\left(\_{}^{30}C\right)_{28}\right)+28\left(\left(\_{}^{30}C\right)_{29}\right)+29\left(\left(\_{}^{30}C\right)_{30}\right)$
Adding we get,
$2S=28\left\{\_{}^{30}C_{0} + \_{}^{30}C_{1} + \_{}^{30}C_{2} + . \ldots . . + \_{}^{30}C_{28} + \_{}^{30}C_{29} + \_{}^{30}C_{30}\right\}$
$\Rightarrow S=14\cdot 2^{30}=\frac{14}{4}2^{32}=\frac{7}{2}\cdot 2^{32}$
$\Rightarrow K=\frac{7}{2}$