Question

Arithmetic Series Algebra 2

Answer 1

The answer is 2400! Yup it is so ya.

Answer 2

`=\displaystyle\sum_(n=1)^(50){(2n)}-\sum_(n=1)^(50){(3)}=2\sum_(n=1)^(50){(n)}-50\cdot 3\\\\=2(50(50+1))/(2)-150=2550-150\\\\=2400`

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The sum of the integers 1 to n is well-known to be n(n+1)/2. Here, we have that sum for n=50.