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Prove it plz :) \[\LARGE{1^2+2^2+3^2+.....+n^2> \frac{n^3}{3}}\]

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@lalaly too:)
do you know standard formula for 1^2+2^2+3^2+.... n^2 =... ?

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Other answers:

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??
no :(
i mean how to expand?
Multiply the terms :) \[(n)(n+1)(2n+1)\]
ok then i gt this \[3n^2+2n^3+n\]
Divide this by 6,
ok then here it is \[\frac{n^2}{2}+\frac{n^3}{3}+\frac{n}{6}.\]
Obviously for positive n \[\large \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} > \frac {n^3}{3} \]
yeah thanx:)
How do you conclude that it is “obvious” for a positive \(n\)? @ash2326
@abhyudaysingh12 got it or not?
for every 'n' it is true actually:)
How would you conclude that?
\(\large \frac{n^2}{2}+\frac{n}{6} >0\) for n>0
^ that way
then add n^3/3 on both sides.
\[n^2 + n > 0 \iff n(n + 1)>0 \iff n +1>0\iff n>-1\]
BUT question says that n>0 so n>-1 is ignored 1,2,.........n see @ParthKohli :)
But you asserted that it's true for ALL \(n\).
sorry for that statement but that i gt in haste;)
lol okay
:D good job;)
to discuss something is good job:)
there are couple of ways you can do it ... few of them are above. Apart from that, you can also try induction.
Mathematical induction: \(1>\frac13\) Suppose this inequality is right for any \(n\). Try to prove it for \(n+1\): \(\sum_{i=1}^{n+1}i^2=\sum_{i=1}^{n}i^2+(n+1)^2>\frac{n^3}3+(n+1)^2=\frac{n^3}3+n^2+2n+1>\) \(>\frac{n^3}3+n^2+n+\frac13=\frac{(n+1)^3}3\) The inequality is proved for \(n+1\). So it is right for all \(n\).

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