Sigma Notation (Summation)

Question 1 of 4

Evaluate

16 \sum n = 5 (5 n - 7)

$\sum_{n=5}^{16}\;(5n-7)$

(546)

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Sum of an Arithmetic Sequence

S_{n} = \frac{n}{2} [a + l]

$S_{\color{#9a00c7}{n}}=\frac{\color{#9a00c7}{n}}{2}[\color{#e65021}{a}+\color{#00880A}{l}]$

First, substitute the values

x = 5 to 16

$x=5 \text(to) 16$ to the expression

$\sum_{\color{#007DDC}{x=5}}^{\color{#007DDC}{16}}\;(5\color{#9a00c7}{n}-7)$	$=$ $=$	$(5 ($ $(5($ $5$ $5$ $) - 7) + (5 ($ $)-7)+(5($ $6$ $6$ $) - 7) + (5 ($ $)-7)+(5($ $7$ $7$ $) - 7) \dots (5 ($ $)-7)…(5($ $16$ $16$ $) - 7)$ $)-7)$
	$=$ $=$	$18$ $18$ $+ 23 + 28 \dots$ $+23+28…$ $73$ $73$

The first term ( $a$ $a$ ) is $18$ $18$ and the last term ( $l$ $l$ ) is $73$ $73$ .

Next, find the number of terms by subtracting the lower value of

x

$x$ from its highest value then adding

1

$1$ .

$n$ $n$	$=$ $=$	$(16 - 5) + 1$ $(16-5)+1$
	$=$ $=$	$12$ $12$

Finally, substitute the known values to the formula

$n$ $n$	$=$ $=$	$12$ $12$
$a$ $a$	$=$ $=$	$18$ $18$
$l$ $l$	$=$ $=$	$73$ $73$

$S_{\color{#9a00c7}{n}}$	$=$ $=$	$\frac{\color{#9a00c7}{n}}{2}[\color{#e65021}{a}+\color{#00880A}{l}]$

$S_{\color{#9a00c7}{12}}$	$=$ $=$	$\frac{\color{#9a00c7}{12}}{2}[\color{#e65021}{18}+\color{#00880A}{73}]$	Substitute known values

	$=$ $=$	$6 [91]$ $6[91]$	Evaluate
	$=$ $=$	$546$ $546$

S_{12} = 546

$S_(12)=546$

Question 2 of 4

Evaluate

10 \sum x = 4 3^{n}

$\sum_{x=4}^{10}\;3^n$

(88533)

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Sum of a Geometric Sequence

S_{n} = \frac{a (r^{n} - 1)}{r - 1}

$S_{\color{#9a00c7}{n}}=\frac{\color{#e65021}{a}(\color{#00880A}{r}^{\color{#9a00c7}{n}}-1)}{\color{#00880A}{r}-1}$

First, substitute the values

x = 4 to 10

$x=4 \text(to) 10$ to the expression

$\sum_{\color{#007DDC}{x=4}}^{\color{#007DDC}{10}}\;3^{\color{#007DDC}{n}}$	$=$ $=$	$3^{\color{#007DDC}{4}}+3^{\color{#007DDC}{5}}+3^{\color{#007DDC}{6}}…3^{\color{#007DDC}{10}}$
	$=$ $=$	$81$ $81$ $+ 243 + 729 \dots 59049$ $+243+729…59 049$

The first term ( $a$ $a$ ) is $81$ $81$ .

Next, find the common ratio ( $r$ $r$ ) by dividing two consecutive terms.

$r$ $r$	$=$ $=$	$3^{5} \div 3^{4}$ $3^5divide3^4$
	$=$ $=$	$3^{5 - 4}$ $3^(5-4)$
	$=$ $=$	$3$ $3$

Next, find the number of terms by subtracting the lower value of

x

$x$ from its highest value then adding

1

$1$ .

$n$ $n$	$=$ $=$	$(10 - 4) + 1$ $(10-4)+1$
	$=$	$7$

Finally, substitute the known values to the formula

$n$	$=$	$7$
$a$	$=$	$81$
$r$	$=$	$3$

$S_{\color{#9a00c7}{n}}$	$=$	$\frac{\color{#e65021}{a}(\color{#00880A}{r}^{\color{#9a00c7}{n}}-1)}{\color{#00880A}{r}-1}$

$S_{\color{#9a00c7}{7}}$	$=$	$\frac{\color{#e65021}{81}(\color{#00880A}{3}^{\color{#9a00c7}{7}}-1)}{\color{#00880A}{3}-1}$	Substitute known values

	$=$	$(81(2186))/2$	Evaluate

	$=$	$(177 066)/2$

	$=$	$88 533$

$S_(7)=88 533$

Question 3 of 4

Find the limiting sum

$\sum_{r=2}^{∞}\;\left(\frac{4}{5}\right)^{r-1}$

$S_∞=$ (4)

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Limiting Sum Formula

$\color{#9a00c7}{S_∞}=\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\text(where) -1$

$<$

$r$

$<$

$1$

Common Ratio Formula

$\color{#00880A}{r}=\frac{U_2}{U_1}=\frac{U_3}{U_2}$

First, substitute the values

$r=2..$ to the expression

$\sum_{\color{#007DDC}{r=2}}^{∞}\;\left(\frac{4}{5}\right)^{r-1}$	$=$	$\left(\frac{4}{5}\right)^{2-1}+\left(\frac{4}{5}\right)^{3-1}\left(\frac{4}{5}\right)^{4-1}…$

	$=$	$4/5$ $+16/25+64/125…$

The first term ( $a$ ) is $4/5$ .

Next, find the common ratio ( $r$ ) by dividing two consecutive terms.

$r$	$=$	$16/25divide4/5$

	$=$	$80/100$

	$=$	$4/5$	Simplify

Finally, substitute the known values to the limiting sum formula

$a$	$=$	$4/5$

$r$	$=$	$4/5$

$\color{#9a00c7}{S_∞}$	$=$	$\frac{\color{#e65021}{a}}{1-\color{#00880A}{r}}$

$\color{#9a00c7}{S_∞}$	$=$	$\frac{\color{#e65021}{\frac{4}{5}}}{1-\color{#00880A}{\frac{4}{5}}}$	Substitute known values

	$=$	$(4/5)/(1/5)$	Evaluate

	$=$	$20/5$

	$=$	$4$

$S_∞=4$

Question 4 of 4

Evaluate

$\sum_{r=3}^{6}\;(-1)^r\;r^2$

(18)

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A sigma notation can be used to get the sum of values without having a common difference or ratio.

Substitute the values

$r=3 \text(to) 6$ to the expression and evaluate

$\sum_{\color{#007DDC}{r=3}}^{\color{#007DDC}{6}}\;(-1)^{\color{#9a00c7}{r}}\;\color{#9a00c7}{r}^2$	$=$	$\left[(-1)^{\color{#007DDC}{3}}\cdot \color{#007DDC}{3}^2\right]+\left[(-1)^{\color{#007DDC}{4}}\cdot \color{#007DDC}{4}^2\right]+\left[(-1)^{\color{#007DDC}{5}}\cdot \color{#007DDC}{5}^2\right]+\left[(-1)^{\color{#007DDC}{6}}\cdot \color{#007DDC}{6}^2\right]$
	$=$	$[(-1)9]+[116]+[(-1)25]+[136]$	Evaluate
	$=$	$-9+16-25+36$
	$=$	$18$

$S_n=18$

Sigma Notation (Summation)

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Quiz summary

Information

1. Question

Hint

Great Work!

Sum of an Arithmetic Sequence

2. Question

Hint

Correct!

Sum of a Geometric Sequence

3. Question

Hint

Keep Going!

Limiting Sum Formula

Common Ratio Formula

4. Question

Hint

Fantastic!

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