If ^{2017}C_0 + ^{2017}C_1 + ^{2017}C_2 + ... | vedclass

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(D) We know that the sum of binomial coefficients is $\sum_{r=0}^{n} {^{n}C_r} = 2^n$.For $n = 2017$,the sum is $\sum_{r=0}^{2017} {^{2017}C_r} = 2^{2

Vedclass · Accepted Answer

$25$

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(D) We know that the sum of binomial coefficients is $\sum_{r=0}^{n} {^{n}C_r} = 2^n$.For $n = 2017$,the sum is $\sum_{r=0}^{2017} {^{2017}C_r} = 2^{2017}$.Since $^{2017}C_r = ^{2017}C_{2017-r}$,we have $\sum_{r=0}^{1008} {^{2017}C_r} = \frac{1}{2} 	imes 2^{2017} = 2^{2016}$.Given $2^{2016} = \lambda^2$,we find $\lambda = \sqrt{2^{2016}} = 2^{1008}$.To find the remainder when $\lambda = 2^{1008}$ is divided by $33$,we use modular arithmetic:$2^5 = 32 \equiv -1 \pmod{33}$.Then $\lambda = 2^{1008} = (2^5)^{201} 	imes 2^3$.$\lambda \equiv (-1)^{201} 	imes 8 \pmod{33}$.$\lambda \equiv -1 	imes 8 \pmod{33} = -8 \pmod{33}$.$-8 + 33 = 25$.Thus,the remainder is $25$.

If $^{2017}C_0 + ^{2017}C_1 + ^{2017}C_2 + ...... + ^{2017}C_{1008} = \lambda^2$ where $\lambda > 0$,then the remainder when $\lambda$ is divided by $33$ is:

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