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 $\sum_{r=0}^{2017} {^{2017}C_r} = 2^{2017}$.Using the property ${^{n}C_r} = {^{n}C_{n-r}}$,we have $\sum_{r=0}^{1008} {^{2017}C_r} = \frac{1}{2} \sum_{r=0}^{2017} {^{2017}C_r} = \frac{2^{2017}}{2} = 2^{2016}$.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 write $2^{1008} = (2^5)^{201} \cdot 2^3 = 32^{201} \cdot 8$.Since $32 \equiv -1 \pmod{33}$,we have $32^{201} \equiv (-1)^{201} \equiv -1 \pmod{33}$.Thus,$\lambda \equiv (-1) \cdot 8 \equiv -8 \pmod{33}$.Since the remainder must be positive,we calculate $-8 + 33 = 25$.

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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