The number (101)^{100}-1 is divisible by | vedclass

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(A) Using the Binomial Theorem, we can write $(101)^{100}-1$ as $(1+100)^{100}-1$. Expanding this using the binomial expansion: $(1+100)^{100}-1 = \le

Vedclass · Accepted Answer

$10^{4}$

Vedclass · Answer

(A) Using the Binomial Theorem, we can write $(101)^{100}-1$ as $(1+100)^{100}-1$. Expanding this using the binomial expansion: $(1+100)^{100}-1 = \left(1 + {}^{100}C_{1}(100) + {}^{100}C_{2}(100)^{2} + {}^{100}C_{3}(100)^{3} + \dots + {}^{100}C_{100}(100)^{100}\right) - 1$. Since ${}^{100}C_{1} = 100$, the expression becomes: $100(100) + {}^{100}C_{2}(100)^{2} + {}^{100}C_{3}(100)^{3} + \dots + {}^{100}C_{100}(100)^{100}$. $= 10^{4} + {}^{100}C_{2}(10^{4}) + {}^{100}C_{3}(10^{6}) + \dots + 10^{200}$. $= 10^{4} \left(1 + {}^{100}C_{2} + {}^{100}C_{3}(10^{2}) + \dots + 10^{196}\right)$. Thus, the expression is divisible by $10^{4}$.

The number $(101)^{100}-1$ is divisible by

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