Question

sigma notation summation help please

Answer 1

\[\sum_{i=1}^{15} i(i-1)^{2}\]

Answer 2

\[\sum_{i=1}^{15}i(i-1)^2=\sum_{i=1}^{15}(i^3-2i^2+i)=\sum_{i=1}^{15}i^3-2\sum_{i=1}^{15}i^2+\sum_{i=1}^{15}i\]
Here are some formulas you'll be using:
\[\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\\
\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}\\
\sum_{i=1}^{n}i^3=\left(\frac{n(n+1)}{2}\right)^2\]

Answer 3

ok let me try that

Answer 4

ok so
[15(15+1)/2]^2
how would u simplify that

Answer 5

\[\left(\frac{15(15+1)}{2}\right)^2=\frac{(15)^2(16)^2}{4}=14,400\]

Answer 6

how about the other 15???

Answer 7

Are you familiar with the properties of exponents? I just applied those and put the fraction into a calculator.

Answer 8

i am but there are 2 15's so i figured you would square both of them

Answer 9

In the numerator:\[(15(15+1))^2=15^2(15+1)^2=15^216^2\]

Answer 10

oooooooooohhh duhhhhhh

Answer 11

thanks

Answer 12

You're welcome

Answer 13

im gonna open another question now can you help

Answer 14

\[\sum_{i=3}^{6}i\]

Answer 15

Use the formula I typed above:
\[\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\]
Be careful, though; the formula is for i=1 to n, and not i=3 to n.

You have to use the following fact:
\[\large \sum_{i=1}^{n}i=\sum_{i=1}^{3}i+\sum_{i=3}^{n}i\]

Answer 16

thats what i dont get
why did yiu go from 1 to 3

Answer 17

in the notebook it says 2 not 3

Answer 18

The sum you're asking about is from 3 to 6. The formula I'm using goes from 1 to 6 (letting n = 6). This means I have to get rid of some terms, so
\[\large\sum_{i=3}^{6}i=\sum_{i=1}^{6}i-\sum_{i=1}^{3}i\]
You typed a 3, too. If it's supposed to be 2, then just switch them around.

Answer 19

so u go from 1 to 3 then from 4 to 6

Answer 20

how about a question like this
\[\sum_{i=51}^{125} i\]

Answer 21

Again, you'd split up the summation like before. And you're right, it should be split up like this:
\[\sum_{i=1}^{6}i=\sum_{i=1}^{2}i+\sum_{i=3}^{6}i\]
Minor mistake on my part.