1 votes 1 votes For positive real numbers $a_1, a_2, \cdots, a_{100}$, let $$p=\sum_{i=1}^{100} a_i \text{ and } q=\sum_{1 \leq i < j \leq 100} a_ia_j.$$ Then $q=\frac{p^2}{2}$ $q^2 \geq \frac{p^2}{2}$ $q< \frac{p^2}{2}$ none of the above Others isi2015-mma summation non-gate + – Arjun asked Sep 23, 2019 recategorized Nov 18, 2019 by Lakshman Bhaiya Arjun 426 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Answer: C $a_i=1$ for $1\le i\le 100$. NastyBall answered Jun 20, 2021 NastyBall comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes C Use this formula, $$\left(\sum_{i==1}^{100} a_i\right)^2=\sum_{i==1}^{100} a_i^2 +2\left(\sum_{1\leq i<j\leq100} a_iaj\right)$$ You can prove this by mathematical induction. Remember that $a_i>0$ for every $i$. KAUNIL answered Mar 10, 2023 KAUNIL comment Share Follow See all 0 reply Please log in or register to add a comment.