The remainder when 7^n - 6n - 50 (n in N) is | vedclass

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(B) We have $7^n = (1 + 6)^n$. By the Binomial Theorem,$7^n = 1 + ^nC_1(6) + ^nC_2(6^2) + ^nC_3(6^3) + \dots + ^nC_n(6^n)$. $7^n = 1 + 6n + 36(^nC_2 +

Vedclass · Accepted Answer

$23$

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(B) We have $7^n = (1 + 6)^n$. By the Binomial Theorem,$7^n = 1 + ^nC_1(6) + ^nC_2(6^2) + ^nC_3(6^3) + \dots + ^nC_n(6^n)$. $7^n = 1 + 6n + 36(^nC_2 + ^nC_3(6) + \dots + ^nC_n(6^{n-2}))$. Let $\lambda = ^nC_2 + ^nC_3(6) + \dots + ^nC_n(6^{n-2})$. Then $7^n = 1 + 6n + 36\lambda$. Rearranging,we get $7^n - 6n = 36\lambda + 1$. Now,subtract $50$ from both sides: $7^n - 6n - 50 = 36\lambda + 1 - 50 = 36\lambda - 49$. We can write $-49$ as $-72 + 23$. $7^n - 6n - 50 = 36\lambda - 72 + 23 = 36(\lambda - 2) + 23$. Let $\mu = \lambda - 2$. Then $7^n - 6n - 50 = 36\mu + 23$. Therefore,when $7^n - 6n - 50$ is divided by $36$,the remainder is $23$.

The remainder when $7^n - 6n - 50$ $(n \in N)$ is divided by $36$ is:

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