How do you find the partial sum of Sigma (n+4)/2 from n=1 to 100?

Feb 11, 2017

${\sum}_{r = 1}^{100} \frac{r + 4}{2} = 2725$

Explanation:

We need the standard formula ${\sum}_{r = 1}^{n} r = \frac{1}{2} n \left(n + 1\right)$

$\therefore {\sum}_{k = 1}^{n} \frac{r + 4}{2} = \frac{1}{2} {\sum}_{k = 1}^{n} \left(r + 4\right)$
$\text{ } = \frac{1}{2} \left\{{\sum}_{k = 1}^{n} r + {\sum}_{k = 1}^{n} 4\right\}$
$\text{ } = \frac{1}{2} \left\{\frac{1}{2} n \left(n + 1\right) + 4 n\right\}$
$\text{ } = \frac{1}{2} \cdot \frac{1}{2} \left\{n \left(n + 1\right) + 8 n\right\}$
$\text{ } = \frac{1}{4} \left\{{n}^{2} + n + 8 n\right\}$
$\text{ } = \frac{1}{4} \left\{{n}^{2} + 9 n\right\}$
$\text{ } = \frac{1}{4} n \left(n + 9\right)$

Put $n = 100 \implies$

$\therefore {\sum}_{r = 1}^{100} \frac{r + 4}{2} = \frac{1}{4} \left(100\right) \left(109\right)$
$\text{ } = 2725$

Feb 11, 2017

${\sum}_{r = 1}^{100} \frac{r + 4}{2} = 2725$

Explanation:

Another approach using the AP formula:

Write out the first few terms to establish the pattern;

${\sum}_{r = 1}^{100} \frac{r + 4}{2} = \frac{5}{2} + \frac{6}{2} + \frac{7}{2} + \ldots \frac{104}{2}$

The terms for an Arithmetic Progression (AP) with $a = \frac{5}{2}$ and $d = \frac{1}{2}$ and there are $100$ terms.

Using the standard AP formula:

${S}_{n} = \frac{1}{2} n \left\{2 a + \left(n - 1\right) d\right\}$

We get;

${S}_{100} = \frac{100}{2} \left\{2 \left(\frac{5}{2}\right) + \left(99\right) \left(\frac{1}{2}\right)\right\}$
$\text{ } = 50 \left\{5 + \frac{99}{2}\right\}$
$\text{ } = 2725$