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The numbers $a_n=6^n-5 n$ for $n=1,2,3, \ldots$ when divided by 25 leave the remainder
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Given, $a_n=6^n-5 n, n=1,2,3, \ldots$
We take; $6^n=(1+5)^n$
Expand with binomial expansion
$6^n={ }^n C_0+{ }^n C_1 5+{ }^n C_2 5^2+{ }^n C_3 5^3+\ldots$
$6^n=1+n \cdot 5+{ }^n C_2 25+{ }^n C_3 5^3+\ldots$
$\left(6^n-5 n\right)=1+25\left\{{ }^n C_2+{ }^n C_3 \cdot 5+\ldots\right\}$
$\left(6^n-5 n\right)=1+25 \cdot k$
where $k=$ positive integer.
Hence, $a_n=6^n-5 n$ divided by 25 and leave the remainder $=1$
We take; $6^n=(1+5)^n$
Expand with binomial expansion
$6^n={ }^n C_0+{ }^n C_1 5+{ }^n C_2 5^2+{ }^n C_3 5^3+\ldots$
$6^n=1+n \cdot 5+{ }^n C_2 25+{ }^n C_3 5^3+\ldots$
$\left(6^n-5 n\right)=1+25\left\{{ }^n C_2+{ }^n C_3 \cdot 5+\ldots\right\}$
$\left(6^n-5 n\right)=1+25 \cdot k$
where $k=$ positive integer.
Hence, $a_n=6^n-5 n$ divided by 25 and leave the remainder $=1$
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