Question

Summation of k=4 to 20 of k^2

Answer 1

\[\sum_{4}^{20} k^2\]

Answer 2

4^2 + 5^2 + .... + 20^2

Answer 3

Is there a way of not expanding the entire thing?

Answer 4

are you in calculus? like calc 2. are you guys doing series.

Answer 5

Yes. I am in Calculus BC.

Answer 6

i haven't taken it in a while so I don't remember if there is a special method to do that.

Answer 7

Hmm. Ok then, thanks anyway. :)

Answer 8

@Algebraic!
@calculusfunctions
@eseidl
@radar
Sorry to bother you guys, but I need to teach my class how to do this tomorrow, so any help would be greatly appreciated. :)

Answer 9

There is a formula to do the sum of the squares of the first n natural numbers. It is:\[\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}\]you could calculate the sum of the first 25 natural numbers, taking n=25. This gives you the sum from n=0 to n=25. To get the sum from 4 to 25 you could subtract the sum you get when n=4 from the first number.

Answer 10

There is a formula for the sum of the first n squares. If you write the question as:\[\sum_{k=4}^{20}k^2=\sum_{k=1}^{20}k^2-\sum_{k=1}^{3}k^2\]you could apply the formula:\[\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}\]

Answer 11

eeer sorry, meant n=20.

Answer 12

@joemath314159 ,
why would the top bound be 3? Wouldn't it be four?

Answer 13

btw, my formula and @joemath314159 are the same thing

Answer 14

Yeah, one if just simplifies more, right?

Answer 15

*is ... simplified

Answer 16

yeah I thought the top bound would be 4 but I could be wrong

Answer 17

Because you want:

4^2+5^2+6^2+...+19^2+20^2

which is the same as:

(1^2+2^2+3^2+4^2+...+19^2+20^2)-(1^2+2^2+3^2).

You only want to subtract the first 3 terms so you dont accidentally get rid of the 4.

Answer 18

OH, ok, that makes sense, thank you! Both of you! :D

Answer 19

no, @joemath314159 must be correct. example, the sum from n=2 to n=3 would be
2^2+3^2=13

or,
\[\frac{27}{3}+\frac{9}{2}+\frac{3}{6}-1=5+9-1=13\]

Answer 20

Yes, he is. :)
Thank you

Answer 21

no problem.