# How do you find the partial sum of Sigma (4.5+0.025j) from j=1 to 200?

Jun 24, 2018

$1402.5$.

#### Explanation:

We know that, sum_(j=1)^nkj=k/2n(n+1), &, sum_(j=1)^nk=kn.

$\therefore {\sum}_{j = 1}^{200} \left(4.5 + 0.025 j\right)$,

$= {\sum}_{j = 1}^{200} 4.5 + {\sum}_{j = 1}^{200} 0.025 j$,

$= \left(4.5\right) \left(200\right) + 0.025 {\sum}_{j = 1}^{200} j$,

$= 900 + 0.025 \left\{\frac{1}{2} \cdot 200 \cdot \left(200 + 1\right)\right\}$,

$= 900 + 502.5$,

$= 1402.5$.