填充第6題的解法參考

文華高中第12題

\(\displaystyle a_1+a_{104}=a_2+a_{103}=a_3+a_{102}=\ldots=a_{52}+a_{53}=\frac{2015}{52}\)
\((a_1b_1+a_2b_2+a_3b_3+\ldots+a_{104}b_{104})+(a_1b_{104}+a_2b_{103}+a_3b_{102}+\ldots+a_{104}b_1)\)
\(=(a_1+a_{104})b_1+(a_2+a_{103})b_2+(a_3+a_{102})b_3+\ldots+(a_{104}+a_1)b_{104}\)
\(\displaystyle =\frac{2015}{52}(b_1+b_2+b_3+\ldots+b_{104})\)
\(\displaystyle =\frac{2015}{52}\times 520\)
\(=20150\)
故所求\(a_1b_{104}+a_2b_{103}+a_3b_{102}+\ldots+a_{104}b_1=150\)