maheshmeghwal9
  • maheshmeghwal9
Prove it plz :) \[\LARGE{1^2+2^2+3^2+.....+n^2> \frac{n^3}{3}}\]
Mathematics
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
maheshmeghwal9
  • maheshmeghwal9
@UnkleRhaukus @hartnn @Hero @satellite73 @amistre64 @Callisto @experimentX Please help:D
maheshmeghwal9
  • maheshmeghwal9
@lalaly too:)
hartnn
  • hartnn
do you know standard formula for 1^2+2^2+3^2+.... n^2 =... ?

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More answers

ash2326
  • ash2326
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??
maheshmeghwal9
  • maheshmeghwal9
no :(
maheshmeghwal9
  • maheshmeghwal9
i mean how to expand?
ash2326
  • ash2326
Multiply the terms :) \[(n)(n+1)(2n+1)\]
maheshmeghwal9
  • maheshmeghwal9
ok then i gt this \[3n^2+2n^3+n\]
ash2326
  • ash2326
Divide this by 6,
maheshmeghwal9
  • maheshmeghwal9
ok then here it is \[\frac{n^2}{2}+\frac{n^3}{3}+\frac{n}{6}.\]
ash2326
  • ash2326
Obviously for positive n \[\large \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} > \frac {n^3}{3} \]
maheshmeghwal9
  • maheshmeghwal9
yeah thanx:)
ParthKohli
  • ParthKohli
How do you conclude that it is “obvious” for a positive \(n\)? @ash2326
maheshmeghwal9
  • maheshmeghwal9
@abhyudaysingh12 got it or not?
maheshmeghwal9
  • maheshmeghwal9
for every 'n' it is true actually:)
ParthKohli
  • ParthKohli
How would you conclude that?
hartnn
  • hartnn
\(\large \frac{n^2}{2}+\frac{n}{6} >0\) for n>0
hartnn
  • hartnn
^ that way
hartnn
  • hartnn
then add n^3/3 on both sides.
ParthKohli
  • ParthKohli
\[n^2 + n > 0 \iff n(n + 1)>0 \iff n +1>0\iff n>-1\]
maheshmeghwal9
  • maheshmeghwal9
BUT question says that n>0 so n>-1 is ignored 1,2,.........n see @ParthKohli :)
ParthKohli
  • ParthKohli
But you asserted that it's true for ALL \(n\).
maheshmeghwal9
  • maheshmeghwal9
sorry for that statement but that i gt in haste;)
ParthKohli
  • ParthKohli
lol okay
maheshmeghwal9
  • maheshmeghwal9
:D good job;)
ParthKohli
  • ParthKohli
?
maheshmeghwal9
  • maheshmeghwal9
to discuss something is good job:)
ParthKohli
  • ParthKohli
?
maheshmeghwal9
  • maheshmeghwal9
?
experimentX
  • experimentX
there are couple of ways you can do it ... few of them are above. Apart from that, you can also try induction.
UnkleRhaukus
  • UnkleRhaukus
|dw:1357174739667:dw|
maheshmeghwal9
  • maheshmeghwal9
|dw:1357869383781:dw|
klimenkov
  • klimenkov
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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