Question

Arithmetic Series Algebra 2

Answer 1

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

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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.

Answer 2

The answer is 2400! Yup it is so ya.