Question

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

• A. 9
• B. 7
• C. 3
• D. 1

Answer 1

Answer: (D)

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 $=1$