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

The numbers $a_n=6^n-5 n$ for $n=1,2,3, \ldots$ when divided by 25 leave the remainder
  1. 9
  2. 7
  3. 3
  4. 1

Solution

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 $=1$

Asked in: AP EAMCET 2010

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