Question

which is greater?
(j=1 to n)summation(square(j))
or
(j=1 to square(n))summation j

Answer 1

is this your question \[\sum\limits_{j=1}^nj^2\]or\[\sum\limits_{j=1}^{n^2}j\]

Answer 2

yes

Answer 3

can you please answer it for me?

Answer 4

i think it depends on \(n\)

Answer 5

i will go with door #2

Answer 6

if i am not mistaken, first one is a polynomial of degree 3 and second one is a polynomial of degree 4, so it is larger
then again i could be wrong

Answer 7

If n=1 they are equal and if n>1 the second one is larger.

Answer 8

ok.you're correct with the degrees, but even the coefficient matters....@satellite73

Answer 9

you have actual formulas for each of these, so i t should be straightforward to compare

Answer 10

how do you prove it?@sauravshakya i knw u cn do it by substituting values,but otherwise...?

Answer 11

there are different degrees in each of them,so u cnt compare...@satellite73

Answer 12

?

Answer 13

yes, i claim a polynomial of degree 4 is larger than a polynomial of degree 3 for sure
after some point, the fourth degree will be necessarily larger than the third degree

Answer 14

if you want to find out at what point the fourth degree surpasses the third degree, i suppose you could be fancy, subtract the third degree from the fourth degree, take the derivative, and see over what interval the function is increasing

Answer 15

\[4*2^{3}>2^{4}\]

Answer 16

yes, that is true, so for some numbers \(4x^3>x^4\) but not in the long run

Answer 17

ok so its only after a certain interval....so in the question either of them could be greater depending on the interval?

Answer 18

if you subtract you get
\[\frac{n^4}{2}-\frac{n^3}{3}-\frac{n}{6}\] and this is increasing on \((1,\infty)\)

Answer 19

yes conceivably the cube could be greater than the fourth degree for some numbers, but not after a certain point.
in your case that point is 1

Answer 20

so door2 will be the correct answer?