Let \(S\) represent the sum of the first \(n\) positive integers. Then, \(S\) and \(n\) satisfy the following equation: \[S = \dfrac{n(n+1)}{2}\]Verify the two equalities below for \(n = 5\) and \(n = 10\):
\(1+2+3+4+5 =\dfrac{5(5+1)}{2} \quad \) and \(\quad 1+2+3+\cdots+9+10 = \dfrac{10(10+1)}{2}\)
Calculate each sum below. You can take the time to add up all of the numbers on a calculator if you'd like, but that won't be much fun! A much better option is to try and find a shortcut using the above formula.
- \(1+2+3+ \cdots + 99 +100\)
- \(19+20+21+\cdots+89+90\)
- \(4+8+12+\cdots+96+100\)
- \(3+7+11+\cdots+83+87\)