Question

Calculate the following series:

Answer 1

\[\sum_{k=1}^{20} 3k^2 +5k\]

Answer 2

HI!!

Answer 3

break it apart first

Answer 4

\[3\sum k^2+5\sum k\] then use the formulas for each

Answer 5

\[\lim_{n \rightarrow \infty}\sum_{k=1}^{n} (4+2k/3\times1/n)\times1/n\]

Answer 6

Hiya! ;)

Answer 7

i can't read the second one

Answer 8

\[\lim_{n \rightarrow \infty}\sum_{k=1}^{n} (1+2k/n)^8 \times1/n\]

Answer 9

One sec, I will edit it.

Answer 10

\[\lim_{n \rightarrow \infty}\sum_{k=1}^{n} (4+\frac{2k}{3})\times\frac{1}{n})\times\frac{1}{n}\] is my guess

Answer 11

some sorta integral right?

Answer 12

turn it to \[\frac{1}{n^2}\sum 4+\frac{2}{3}\sum k\]

Answer 13

actually \[\frac{1}{n^2}\left(\sum_{k=1}^n 4+\frac{2}{3}\sum_{k=1}^n k\right)\]

Answer 14

unless am reading it wrong, which is possible

Answer 15

\[\lim_{n \rightarrow \infty}\sum_{k=1}^{n} (1+2k/n)^8 \times1/n\]

are you supposed to write this as an integral and evaluate?

Answer 16

Just says evaluate, your thought is as good as mine.

Answer 17

Not to be blunt, I just don't know