We often encounter triple sums such as $$ S_1 = \sum_i \sum_j \sum_k (1 \le i \ne j \ne k \le n) ~~f_i f_j f_k $$ $$ S_2 = \sum_i \sum_j \sum_k (1 \le i < j < k < n) ~~f_i f_j f_k $$ $$ S_3 = \sum_i \sum_j \sum_k (1 \le i \le j \le k \le n) ~~f_i f_j f_k~. $$
I would like to know if $ S_1 , S_2 , S_3 $ can be expressed in terms of $\sum f_i~,\sum f^2 _i ~\mbox{and}~ \sum f^3 _i ~ . $ Please help.
Yes, due to the the interesting identity: $$(a+b+c)^3=3(a^2+b^2+c^2)(a+b+c)-2(a^3+b^3+c^3)+6abc.$$ $S_1,S_2,S_3$ in your question can be expressed as: $$S_1=\left(\sum_{i}f_i\right)^3-3 \left(\sum_i f^2_i \right) \left(\sum_i f_i\right)+2 \left(\sum_i f^3_i\right)$$
$$ S_2 = \frac{1}{6}\left[\left(\sum_{i}f_i\right)^3-3 \left(\sum_i f^2_i \right) \left(\sum_i f_i\right)+2 \left(\sum_i f^3_i\right)\right]$$
$$ S_3 = \frac{1}{6}\left[\left(\sum_{i}f_i\right)^3 -3\left(\sum_i f^2_i \right) \left(\sum_i f_i\right) + 8\left(\sum_i f^3_i\right)\right] $$