Question

Evaluate the sum of.... drawn inside! thanks!! :D

Answer 1

not sure what it means when there's nothing above the sigma! :/

Answer 2

are you sure the question is \[\huge\color{blue}{ \sum_{k=1}^{~} 13k^{2}} \]

Answer 3

sure there is nothing above the sigma ?

Answer 4

yes :/

not sure what that means!! :/

Answer 5

well, when there is a number 6 above the sigma that means that there are six terms, if ∞ is above the sigma, then it is an infinite series (that either diverges or converges), but when there is nothing above the sigma, then it is an unknown number of terms, and therefore I son't think the sum can be determined.

Answer 6

ahh okay i see.. thank you!!

Answer 7

Anytime

Answer 8

if we were to put a random number on top, like 4, how would we go about solving?

Answer 9

or any number really... haha

Answer 10

then we would plug in:

1 for k, (into 13k² ) to find the first term
2 for k, to find the second term
3 for k, to find the third term
4 for k, to find the forth term

and the sum of the four terms would be the answer.

Answer 11

okay

so like this?

13(1)^2 + 13(2)^2 + 13(3)^2 + 13(4)^2 ?

and we get 390 ?

Answer 12

Yes, just like that.
\[13(1)^{2} + 13(2)^{2} + 13(3)^{2} + 13(4)^{2} \]\[13 + (13 \times 4) + (13 \times 9)+ (13 \times 16)\]\[13 + 52+ 117+ 208\]\[390~~.\]

Answer 13

Since sigma is the sum we can just say
\[13(1+4+9+16+25+36....)\]
for THE SUM of whatever number of term

Answer 14

ahh okay awesome! thank you!! :D

i have two more like this, but they actually have numbers over the sigma this time (!!!)... haha could you possibly help me with them? if so, should i post here? or a new post? :)

Answer 15

I am not good these, but I can try ;)

Answer 16

okay awesome! i'll start a new post :)