The numbers a_n = 6^n - 5n for n = 1, 2, 3, l | vedclass

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(D) Given,$a_n = 6^n - 5n$ for $n = 1, 2, 3, \ldots$ We can write $6^n$ as $(1 + 5)^n$. Using the binomial expansion: $6^n = (1 + 5)^n = {^nC_0} + {^n

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$1$

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(D) Given,$a_n = 6^n - 5n$ for $n = 1, 2, 3, \ldots$ We can write $6^n$ as $(1 + 5)^n$. Using the binomial expansion: $6^n = (1 + 5)^n = {^nC_0} + {^nC_1}(5) + {^nC_2}(5^2) + {^nC_3}(5^3) + \ldots + {^nC_n}(5^n)$ $6^n = 1 + 5n + 25({^nC_2} + {^nC_3}(5) + \ldots + {^nC_n}(5^{n-2}))$ Now,subtract $5n$ from both sides: $a_n = 6^n - 5n = 1 + 25({^nC_2} + 5{^nC_3} + \ldots + 5^{n-2}{^nC_n})$ Let $k = {^nC_2} + 5{^nC_3} + \ldots + 5^{n-2}{^nC_n}$,where $k$ is an integer for $n \geq 2$. Thus,$a_n = 1 + 25k$. For $n = 1$,$a_1 = 6^1 - 5(1) = 1$. In both cases,$a_n$ divided by $25$ leaves a remainder of $1$.

The numbers $a_n = 6^n - 5n$ for $n = 1, 2, 3, \ldots$ when divided by $25$ leave the remainder:

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