maheshmeghwal9
Prove it plz :)
\[\LARGE{1^2+2^2+3^2+.....+n^2> \frac{n^3}{3}}\]
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maheshmeghwal9
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@UnkleRhaukus @hartnn @Hero @satellite73 @amistre64 @Callisto @experimentX Please help:D
maheshmeghwal9
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@lalaly too:)
hartnn
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do you know standard formula for 1^2+2^2+3^2+.... n^2 =... ?
ash2326
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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
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no :(
maheshmeghwal9
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i mean how to expand?
ash2326
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Multiply the terms :)
\[(n)(n+1)(2n+1)\]
maheshmeghwal9
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ok then i gt this
\[3n^2+2n^3+n\]
ash2326
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Divide this by 6,
maheshmeghwal9
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ok then here it is \[\frac{n^2}{2}+\frac{n^3}{3}+\frac{n}{6}.\]
ash2326
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Obviously for positive n
\[\large \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} > \frac {n^3}{3} \]
maheshmeghwal9
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yeah thanx:)
ParthKohli
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How do you conclude that it is “obvious” for a positive \(n\)? @ash2326
maheshmeghwal9
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@abhyudaysingh12 got it or not?
maheshmeghwal9
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for every 'n' it is true actually:)
ParthKohli
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How would you conclude that?
hartnn
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\(\large \frac{n^2}{2}+\frac{n}{6} >0\)
for n>0
hartnn
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^ that way
hartnn
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then add n^3/3 on both sides.
ParthKohli
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\[n^2 + n > 0 \iff n(n + 1)>0 \iff n +1>0\iff n>-1\]
maheshmeghwal9
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BUT question says that n>0 so n>-1 is ignored
1,2,.........n
see @ParthKohli :)
ParthKohli
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But you asserted that it's true for ALL \(n\).
maheshmeghwal9
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sorry for that statement
but that i gt in haste;)
ParthKohli
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lol okay
maheshmeghwal9
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:D
good job;)
ParthKohli
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?
maheshmeghwal9
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to discuss something is good job:)
ParthKohli
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?
maheshmeghwal9
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?
experimentX
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there are couple of ways you can do it ... few of them are above.
Apart from that, you can also try induction.
UnkleRhaukus
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|dw:1357174739667:dw|
maheshmeghwal9
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|dw:1357869383781:dw|
klimenkov
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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 \(n\).