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

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maheshmeghwal9Best ResponseYou've already chosen the best response.1
@UnkleRhaukus @hartnn @Hero @satellite73 @amistre64 @Callisto @experimentX Please help:D
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
do you know standard formula for 1^2+2^2+3^2+.... n^2 =... ?
 one year ago

ash2326Best ResponseYou've already chosen the best response.5
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??
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.1
i mean how to expand?
 one year ago

ash2326Best ResponseYou've already chosen the best response.5
Multiply the terms :) \[(n)(n+1)(2n+1)\]
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.1
ok then i gt this \[3n^2+2n^3+n\]
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.1
ok then here it is \[\frac{n^2}{2}+\frac{n^3}{3}+\frac{n}{6}.\]
 one year ago

ash2326Best ResponseYou've already chosen the best response.5
Obviously for positive n \[\large \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} > \frac {n^3}{3} \]
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
How do you conclude that it is “obvious” for a positive \(n\)? @ash2326
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.1
@abhyudaysingh12 got it or not?
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.1
for every 'n' it is true actually:)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
How would you conclude that?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
\(\large \frac{n^2}{2}+\frac{n}{6} >0\) for n>0
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
then add n^3/3 on both sides.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
\[n^2 + n > 0 \iff n(n + 1)>0 \iff n +1>0\iff n>1\]
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.1
BUT question says that n>0 so n>1 is ignored 1,2,.........n see @ParthKohli :)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
But you asserted that it's true for ALL \(n\).
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.1
sorry for that statement but that i gt in haste;)
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.1
to discuss something is good job:)
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
there are couple of ways you can do it ... few of them are above. Apart from that, you can also try induction.
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
dw:1357174739667:dw
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.1
dw:1357869383781:dw
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
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 \(n\).
 one year ago
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