## maheshmeghwal9 3 years ago Prove it plz :) $\LARGE{1^2+2^2+3^2+.....+n^2> \frac{n^3}{3}}$

1. maheshmeghwal9

2. maheshmeghwal9

@lalaly too:)

3. hartnn

do you know standard formula for 1^2+2^2+3^2+.... n^2 =... ?

4. 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??

5. maheshmeghwal9

no :(

6. maheshmeghwal9

i mean how to expand?

7. ash2326

Multiply the terms :) $(n)(n+1)(2n+1)$

8. maheshmeghwal9

ok then i gt this $3n^2+2n^3+n$

9. ash2326

Divide this by 6,

10. maheshmeghwal9

ok then here it is $\frac{n^2}{2}+\frac{n^3}{3}+\frac{n}{6}.$

11. ash2326

Obviously for positive n $\large \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} > \frac {n^3}{3}$

12. maheshmeghwal9

yeah thanx:)

13. ParthKohli

How do you conclude that it is “obvious” for a positive $$n$$? @ash2326

14. maheshmeghwal9

@abhyudaysingh12 got it or not?

15. maheshmeghwal9

for every 'n' it is true actually:)

16. ParthKohli

How would you conclude that?

17. hartnn

$$\large \frac{n^2}{2}+\frac{n}{6} >0$$ for n>0

18. hartnn

^ that way

19. hartnn

then add n^3/3 on both sides.

20. ParthKohli

$n^2 + n > 0 \iff n(n + 1)>0 \iff n +1>0\iff n>-1$

21. maheshmeghwal9

BUT question says that n>0 so n>-1 is ignored 1,2,.........n see @ParthKohli :)

22. ParthKohli

But you asserted that it's true for ALL $$n$$.

23. maheshmeghwal9

sorry for that statement but that i gt in haste;)

24. ParthKohli

lol okay

25. maheshmeghwal9

:D good job;)

26. ParthKohli

?

27. maheshmeghwal9

to discuss something is good job:)

28. ParthKohli

?

29. maheshmeghwal9

?

30. experimentX

there are couple of ways you can do it ... few of them are above. Apart from that, you can also try induction.

31. UnkleRhaukus

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32. maheshmeghwal9

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33. 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 $$n$$.