Question

\[\delta_{ij}\sum_{i=12}^{12}\prod_{k=j}^{k}j^2\]

Answer 1

what do you mean by
k=j to k ?

Answer 2

j=k*

Answer 3

and i = 12 to 12 ? :O

Answer 4

thats right

Answer 5

so as I see it:
there is not really summing nor multiplying ..
it looks like k = 12 and that is it.
so 12^2 = 144 (i = j otherwise the delta is zero)

Answer 6

you got it !

Answer 7

:)

Answer 8

\(\sum \limits_{i=12}^{12}k^2=k^2\)
right ?

Answer 9

√

Answer 10

so, \(\delta_{ij}k^2=0\)
so, 0 is final answer ?? just confirming...

Answer 11

notice i made an error when i wrote out the question the first time i put k=j when i meant j=k

Answer 12

\(\delta_{ij}\sum_{i=12}^{12}\prod_{j=k}^{k}j^2=\delta_{ij}\sum_{i=12}^{12}k^2=\delta_{ij}k^2=0\)
?

Answer 13

hmm i guess i should have left it as k=j

Answer 14

no the variable is j, so it must be
j= something to something

Answer 15

i don't get how its 144

Answer 16

@Coolsector , how is k=12 ??

Answer 17

the variable doesn't have to appear in the argument