Question

Calculate the following series:

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

Yes.

Answer 5

Now use the results:\[\sum_{k=1}^n k = \frac{n(n+1)}{2}\]\[\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}\]

Answer 6

3(8610)+5(210)

Answer 7

= 26880

Answer 8

I guess so.

Answer 9

For the second one, just simplify the whole thing and then see what happens when \(n \to \infty\) afterwards

Answer 10

So take the limit after doing the same process to the series?

Answer 11

Yeah, don't worry about the limit. We'll handle that later. Just simplify the expression first.

Answer 12

First \[\lim_{n \rightarrow \infty}\sum_{k=1}^{20} (\frac{ 4 }{ n }+\frac{ 2k^3 }{ n }×\frac{ 1 }{ n^2 })\]

Answer 13

Where did the 20 come in the picture?

Answer 14

Actually I don't even need to do that. \[4\sum_{k=1}^{n}\frac{ 1 }{ n } + 2\sum_{k=1}^{n}\frac{ k^3 }{ n } + \sum_{k=1}^{n}\frac{ 1 }{ n^2 }\]

Answer 15

Sorry, my mistake

Answer 16

Yes, also note that \(n\) is a constant in the summation.

Answer 17

Noted.

Answer 18

lol ok, now calculate the rest. don't think about the limit until you're done.

Answer 19

Ok.
((n^2+n)/2) + ((4n^4 + 12n^2)/4) + ((2n^3 + n^2 +3n+n)/6)

Answer 20

Sorry that took so long.

Answer 21

Simplified that would be:

Answer 22

oh what

Answer 23

\[\frac{ n^2 + n}{ 2 }+\frac{ 4n^4 +12n^2}{ 4 }+\frac{ 2n^3 +n^2+4n}{ 6 } \rightarrow \frac{ 6n^2 + 6n }{ 12 }+ \frac{ 12n^4 +36n^2}{ 12 } + \frac{ 4n^3 + 2n^2+8n}{ 12 }\]

Answer 24

my god no

Answer 25

lol

Answer 26

\[\frac{4}{n}+\frac{2k}{3n^2}\]is this your summand?

Answer 27

Sure

Answer 28

Did anyone calculate the sum to be 9660 ?

Answer 29

yes 9660 is correct @robtobey