# Compute the following two sums: (a) S_a =Sigma_(k=1)^20(2 − 3k + 2k^2), (b)S_b =Sigma_(k=10)^50k?

Jun 1, 2016

We will use the following formulae:

• ${\sum}_{k = 1}^{n} c {X}_{k} = c {\sum}_{k = 1}^{n} {X}_{k}$
• ${\sum}_{k = 1}^{n} \left({X}_{k} \pm {Y}_{k}\right) = {\sum}_{k = 1}^{n} {X}_{k} \pm {\sum}_{k = 1}^{n} {Y}_{k}$
• ${\sum}_{k = 1}^{n} c = c n$
• ${\sum}_{k = 1}^{n} k = \frac{n \left(n + 1\right)}{2}$
• ${\sum}_{k = 1}^{n} {k}^{2} = \frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}$

(a)

${\sum}_{k = 1}^{20} \left(2 - 3 k + 2 {k}^{2}\right) = {\sum}_{k = 1}^{20} 2 - 3 {\sum}_{k = 1}^{20} k + 2 {\sum}_{k = 1}^{20} {k}^{2}$

Here, $20 = n$ in all instances:

$= 2 \cdot 20 - 3 \left(\frac{\left(20\right) \left(20 + 1\right)}{2}\right) + 2 \left(\frac{20 \left(20 + 1\right) \left(2 \cdot 20 + 1\right)}{6}\right)$

$= 40 - 630 + 5740 = 5150$

(b)

Note that:

${\overbrace{{\sum}_{k = 10}^{50} k}}^{\text{10,11...49,50")=overbrace(sum_(k=1)^50k)^("1,2...49,50")-overbrace(sum_(k=1)^9k)^("1,2...8,9}}$

So, this then equals

$\frac{\left(50\right) \left(50 + 1\right)}{2} - \frac{\left(9\right) \left(9 + 1\right)}{2} = 1275 - 45 = 1230$