Question

Can someone PLEASE help me with Sigma Notation??

Answer 1

lol srry im only in 10th grade lol im not learning that yet lol

Answer 2

i would adjust the index, let n = i+2

Answer 3

@amistre64 whys that?

Answer 4

its easier to assess when the index starts at 1

Answer 5

\[\large\sum_{n=a}^{b}f(n)\]
let n=i+a-1
\[\large\sum_{i+a-1=a}^{b}f(i+a-1)\]
\[\large\sum_{i=a-(a-1)}^{b-(a-1)}f(i+a-1)\]

Answer 6

this is confused, jeeze :(

Answer 7

in this case ....

\[\large\sum_{n=3}^{12}~-5n-1\]
\[\large\sum_{n=3-2}^{12-2}~-5(n+2)-1\]
\[\large\sum_{n=1}^{10}~-5n-10-1\]
\[\large\sum_{n=1}^{10}~-5n-11\]

Answer 8

alright, that makes sense. what next?

Answer 9

well, we can brute it ...

-5(1) - 11
-5(2) - 11
-5(3) - 11
-5(4) - 11
......
-5(10) - 11
-----------
-5(1+2+3+4+..+10)
-11(10)

Answer 10

if you know how to add the numbers from 1 to 10 thats the hardest issue i believe

Answer 11

alright

Answer 12

I got the answer!
its -385

Answer 13

if we do the summation notations ... it would look a little like this:
\[\sum_{n=3}^{12} -5n-1\]
\[\left(\sum_{n=1}^{12} -5n-1\right)-\left(\sum_{n=1}^{2} -5n-1\right)\]
\[\left(-5\sum_{n=1}^{12} n-\sum_{n=1}^{12}1\right)-\left(-5\sum_{n=1}^{2}n-\sum_{n=1}^{2}1\right)\]
\[\left(-5\sum_{n=1}^{12} n-12\right)-\left(-5\sum_{n=1}^{2}n-12\right)\]
\[\left(-5\frac{12(13)}{2}-12\right)-\left(-5\frac{2(3)}{2}-12\right)\]

Answer 14

-5(6)(13) + 5(3)

5 (-(6)(13) + 3)

5 (-78 + 3)

5 (-75) = -375

Answer 15

lol, typo ... -385 yes

Answer 16

thank you so much! (: @amistre64

Answer 17

good luck