Question

A question pertaining to Sigma Notation! When evaluating a sum, if the part to the right of the sigma is "k^3 - 45", can it be analyzed, using summation rules, as follows? ->
"((n(n+1))/2)^2 + n(45)" ??

Any and all help is greatly appreciated! :)

Answer 1

yup !

Answer 2

\(\large \sum \limits_{k=1}^n (k^3 - 45)\)

Answer 3

Assuming it's a finite sum, sure. You have to be careful if you're summing an infinite series, though...

Answer 4

Thank you so much! I just wanted to double check my conceptual understanding there! :)

Answer 5

Yes, there is a finite limit:)

Answer 6

\(\large \sum \limits_{k=1}^n (k^3 - 45) = \sum \limits_{k=1}^n (k^3 ) - \sum \limits_{k=1}^n ( 45) \)

Answer 7

u wlc :)

Answer 8

So, say n was 5, and k was 1.... The second part is still just going to be found as 45? Not 5(45)?

Answer 9

\(\large \sum \limits_{k=1}^5 (k^3 - 45) = \sum \limits_{k=1}^5 (k^3 ) - \sum \limits_{k=1}^5 ( 45) \)


\(\large \qquad \qquad~~~~~ = \sum \limits_{k=1}^5 (k^3 ) - 45\sum \limits_{k=1}^5 ( 1) \)

Answer 10

... And the second part (aside from the coefficient of 45 now), is equal to five?

Answer 11

yes !

Answer 12

\(\large \sum \limits_{k=1}^5 ( 1) = 1 + 1 + 1 + 1 + 1 = 5\)

Answer 13

...how? .. If there 's no place to plug in values of k? Does this method/ rule hold true for all sigmas with an integer to the right of the sigma, but no variable?

Answer 14

I see what happened, I just don't get how that... Happens, If there is not a place for k values...

Answer 15

\(\large \sum \limits_{k=1}^5 ( 1) = 1 + 1 + 1 + 1 + 1 = 5\)



is same as



\(\large \sum \limits_{k=1}^5 ( k^0 1) = (1^0)1 + (2^0)1 + (3^0)1 + (4^0)1 + (5^0)1 = 5\)

Answer 16

..... Ahhh, lightbulb!!! :D (the world needs more people like you ganeshie8)

Answer 17

the definition of SUM notation is :-
u simply take the sums of the right side thing, each time changing the value of variable \(k\)

Answer 18

lol, okay so below just some examples :-


\(\large \sum \limits_{k=1}^5 ( k) = 1 + 2 + 3 +4 + 5 \)

Answer 19

\(\large \sum \limits_{k=1}^5 ( k^2) = 1^2 + 2^2 + 3^2 +4^2 + 5^2 \)

Answer 20

\( \sum \limits_{k=1}^5 ( k^2(k-k^{300})) = \\ \\ 1^2(1-1^{300}) + 2^2(2-2^{300}) + 3^2(3-3^{300}) +4^2(4-4^{300}) + 5^2(5-5^{300}) \)

Answer 21

\(\sum \) is just a notational thing,
how u evaluate the sums is entirely a different thing....
u could use sum of first n terms formula or sum of squares of first n terms formula... or watever u r familiar with

Answer 22

More complex, but I still follow:) it was mainly the lack of a K to the right of the sign that was confusing me:) but I feel totally confident in how to navigate it now! Thank you again! :)

Answer 23

np :)