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

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(A) Using the Binomial Theorem,we can write $(101)^{100} = (1 + 100)^{100}$. Expanding this using the formula $(1+x)^n = 1 + nC_1 x + nC_2 x^2 + nC_3

Vedclass · Accepted Answer

$10^4$

Vedclass · Answer

(A) Using the Binomial Theorem,we can write $(101)^{100} = (1 + 100)^{100}$. Expanding this using the formula $(1+x)^n = 1 + nC_1 x + nC_2 x^2 + nC_3 x^3 + \ldots + nC_n x^n$: $(1 + 100)^{100} = 1 + {}^{100}C_1(100) + {}^{100}C_2(100)^2 + {}^{100}C_3(100)^3 + \ldots + {}^{100}C_{100}(100)^{100}$. Subtracting $1$ from both sides: $(101)^{100} - 1 = {}^{100}C_1(100) + {}^{100}C_2(100)^2 + {}^{100}C_3(100)^3 + \ldots + {}^{100}C_{100}(100)^{100}$. Since ${}^{100}C_1 = 100$,the first term is $100 	imes 100 = 100^2 = 10^4$. Thus,$(101)^{100} - 1 = 10^4 + {}^{100}C_2(100)^2 + \ldots = 10^4 [1 + {}^{100}C_2(100) + \ldots]$. Therefore,the number is divisible by $10^4$.

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

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