Question

tính A=3+5+7+.....+79+100

Answer 1

\(A=3+5+7+.....+97+99\)

\(\Leftrightarrow A=\dfrac{\left(99+3\right).\left[\left(99-3\right):2+1\right]}{2}\)

\(\Leftrightarrow A=\dfrac{102.\left(96:2+1\right)}{2}\)

\(\Leftrightarrow A=\dfrac{102.\left(48+1\right)}{2}\)

\(\Leftrightarrow A=\dfrac{102.49}{2}\)

\(\Leftrightarrow A=\dfrac{4998}{2}\)

\(\Leftrightarrow A=2499\)

Answer 2

Số số hạng từ \(3\rightarrow79\) là:

\(\left(79-3\right):2+1=39\) ( số hạng )

\(\Rightarrow A=\dfrac{\left(79+3\right).39}{2}+100=1699\)