Sigma notation

[Dionysus1]

I need help with sigma notation
How do I write this in sigma notation?
3+12+21+30+39+48

 

[pka]

I need help with sigma notation
How do I write this in sigma notation?
3+12+21+30+39+48

\sum\limits_{k = 0}^5 {\left( {3 + 9\cdot k} \right)} gives \(\sum\limits_{k = 0}^5 {\left( {3 + 9\cdot k} \right)} \)

 

[Deleted member 4993]

I need help with sigma notation
How do I write this in sigma notation?
3+12+21+30+39+48

First you should note that the given series is in arithmetic progression

a(1) = 3 + 9 * (1 - 1)

a(2) = 3 + 9 * (2 - 1)

a(n) = 3 + ??

Continue......

 

[Deleted member 4993]

\sum\limits_{k = 0}^5 {\left( {3 + 9\cdot k} \right)} gives \(\sum\limits_{k = 0}^5 {\left( {3 + 9\cdot k} \right)} \)

What happened to the "discovery" method?

 

[Steven G]

Here is how I think about it. The numbers all differ by 9, so I think of the 9 times table. 9*1, 9*2, 9*3, 9*4, 9*5 and 9*6
The problem is 9*1= 9 and the first number is 3. Well 9-6=3. In fact, if I subtract 6 from each number on the 9 times table I get the values needed.
[math]\sum\limits_{k = 1}^6 {\left( {9\cdot k - 6} \right)}[/math]

 

[JeffM]

And jomo's way leads very naturally to

[MATH]\sum_{k=1}^n (9k - 6) = - 6n + 9 * \sum_{k=1}^n k = \dfrac{-12n}{2} + \dfrac{9n^2 + 9n}{2} = \frac{3}{2} * (3n^2 - n).[/MATH]
Let's see how that works

[MATH]n = 1 \implies \frac{3}{2} * (3n^2 - n) = \frac{3}{2} * (3 - 1) = 3. \ \checkmark[/MATH]
[MATH]n = 2 \implies \frac{3}{2} * (3n^2 - n) = \frac{3}{2} * (12 - 2) = 15 = 3 + 12.\ \checkmark[/MATH]
[MATH]n = 3 \implies \frac{3}{2} * (3n^2 - n) = \frac{3}{2} * (27 - 3) = 36 = 15 + 21.\ \checkmark[/MATH]
[MATH]n = 4 \implies \frac{3}{2} * (3n^2 - n) = \frac{3}{2} * (48 - 4) = 66 = 36 + 30.\ \checkmark[/MATH]
[MATH]n = 5 \implies \frac{3}{2} * (3n^2 - n) = \frac{3}{2} * (75 - 5) = 105 = 66 + 39.\ \checkmark[/MATH]
[MATH]n = 6 \implies \frac{3}{2} * (3n^2 - n) = \frac{3}{2} * (108 - 6) = 153 = 105 + 48.\ \checkmark[/MATH]