Mathematics - Permutation & Combination Question with Solution | TestHub

Sum of even divisors of $2^{{\alpha }_1}\cdot p_1{}^{{\alpha }_2}\cdot p_2{}^{{\alpha }_3}-----p_k{}^{{\alpha }_4+1}$ is $\text{ 2. }\frac{2^{{\alpha }_1}-1}{2-1}\times \frac{p_1^{{\alpha }_2+1}-1}{p_1-1}\times \frac{p_2^{{\alpha }_2+1}-1}{p_2-1}\times -----\frac{p_k^{{\alpha }_k+1}-1}{p_k-1}$