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maheshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.1@UnkleRhaukus @hartnn @Hero @satellite73 @amistre64 @Callisto @experimentX Please help:D

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

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.5Sum 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
 2 years ago
Best ResponseYou've already chosen the best response.1i mean how to expand?

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.5Multiply the terms :) \[(n)(n+1)(2n+1)\]

maheshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.1ok then i gt this \[3n^2+2n^3+n\]

maheshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.1ok then here it is \[\frac{n^2}{2}+\frac{n^3}{3}+\frac{n}{6}.\]

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.5Obviously for positive n \[\large \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} > \frac {n^3}{3} \]

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

maheshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.1@abhyudaysingh12 got it or not?

maheshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.1for every 'n' it is true actually:)

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0How would you conclude that?

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

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1then add n^3/3 on both sides.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0\[n^2 + n > 0 \iff n(n + 1)>0 \iff n +1>0\iff n>1\]

maheshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.1BUT question says that n>0 so n>1 is ignored 1,2,.........n see @ParthKohli :)

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0But you asserted that it's true for ALL \(n\).

maheshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.1sorry for that statement but that i gt in haste;)

maheshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.1to discuss something is good job:)

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

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1357174739667:dw

maheshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1357869383781:dw

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.0Mathematical 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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