Question

The sum of divisors of $a^{l_1}{b}^{l_2}{c}^{l_3}...k^{l_n}$ is

• A. $a^{l_1}{b}^{l_2}{c}^{l_3}...k^{l_n}$
• B. ${(a+b+c+...+k)}^{l_1+l_2+l_3+...l_n}$
• C. ${(a+1)}^{l_1}{(b+1)}^{l_2}...{(k+1)}^{l_n}$
• D. None of these

Answer 1

The correct option is D None of these

Note that the divisor of the number $n=a^{l_1}{b}^{l_2}\cdots k^{l_n}$ is of the form $n=a^{m_1}{b}^{m_2}\cdots k^{m_n}$ where $0\le m_i\le l_i$ for all $i=1,2,\cdots ,n$
Consider the expression $\left(1+a+a^2+\cdots +a^{l_1}\right)\left(1+b+b^2+\cdots +b^{l_2}\right)\cdots \cdots \left(1+k+k^2+\cdots +k^{l_n}\right)$ (1)
Each term of the expanded product (1) is the divisor of $n$.and hence (1) is sum of the divisor of $n$.
$\left(1+a+a^2+\cdots +a^{l_1}\right)\left(1+b+b^2+\cdots +b^{l_2}\right)\cdots \cdots \left(1+k+k^2+\cdots +k^{l_n}\right)$
$=\left(\frac{a^{l_1+1}-1}{a-1}\right)\left(\frac{b^{l_2+1}-1}{b-1}\right)\cdots \cdots \left(\frac{k^{l_n+1}-1}{k-1}\right)$