Question

Prove it plz :)

\[\LARGE{1^2+2^2+3^2+.....+n^2> \frac{n^3}{3}}\]

Answer 1

@UnkleRhaukus @hartnn @Hero @satellite73 @amistre64 @Callisto @experimentX Please help:D

Answer 2

@lalaly too:)

Answer 3

do you know standard formula for 1^2+2^2+3^2+.... n^2 =... ?

Answer 4

Sum of \(\large 1^2+2^2+3^2...n^2=\frac{(n)(n+1)(2n+1)}{6}\)
If you expand this, you can prove it easily. Can you do that @maheshmeghwal9??

Answer 5

no :(

Answer 6

i mean how to expand?

Answer 7

Multiply the terms :)
\[(n)(n+1)(2n+1)\]

Answer 8

ok then i gt this

\[3n^2+2n^3+n\]

Answer 9

Divide this by 6,

Answer 10

ok then here it is \[\frac{n^2}{2}+\frac{n^3}{3}+\frac{n}{6}.\]

Answer 11

Obviously for positive n
\[\large \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} > \frac {n^3}{3} \]

Answer 12

yeah thanx:)

Answer 13

How do you conclude that it is “obvious” for a positive \(n\)? @ash2326

Answer 14

@abhyudaysingh12 got it or not?

Answer 15

for every 'n' it is true actually:)

Answer 16

How would you conclude that?

Answer 17

\(\large \frac{n^2}{2}+\frac{n}{6} >0\)
for n>0

Answer 18

^ that way

Answer 19

then add n^3/3 on both sides.

Answer 20

\[n^2 + n > 0 \iff n(n + 1)>0 \iff n +1>0\iff n>-1\]

Answer 21

BUT question says that n>0 so n>-1 is ignored
1,2,.........n
see @ParthKohli :)

Answer 22

But you asserted that it's true for ALL \(n\).

Answer 23

sorry for that statement
but that i gt in haste;)

Answer 24

lol okay

Answer 25

:D
good job;)

Answer 26

?

Answer 27

to discuss something is good job:)

Answer 28

?

Answer 29

?

Answer 30

there are couple of ways you can do it ... few of them are above.
Apart from that, you can also try induction.