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

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

ash2326
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1i mean how to expand?

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

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

maheshmeghwal9
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0How do you conclude that it is “obvious” for a positive \(n\)? @ash2326

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

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

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

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

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

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

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

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

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

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

klimenkov
 one year 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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