Question

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

• A. 9
• B. 7
• C. 3
• D. 1

Answer 1

Answer: (D)

Given, $a_n=6^n-5n,n=1,2,3,\dots$
We take; $6^n=(1+5{)}^n$
Expand with binomial expansion
$6^n={}^n{C}_0+{}^n{C}_{1}5+{}^n{C}_2{5}^2+{}^n{C}_3{5}^3+\dots$
$6^n=1+n\cdot 5+{}^n{C}_{2}25+{}^n{C}_3{5}^3+\dots$
$\left(6^n-5n\right)=1+25\left\{{}^n{C}_2+{}^n{C}_3\cdot 5+\dots \right\}$
$\left(6^n-5n\right)=1+25\cdot k$
where $k=$ positive integer.
Hence, $a_n=6^n-5n$ divided by 25 and leave the remainder $=1$