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maheshmeghwal9

Prove it plz :) \[\LARGE{1^2+2^2+3^2+.....+n^2> \frac{n^3}{3}}\]

  • one year ago
  • one year ago

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  1. maheshmeghwal9
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    @UnkleRhaukus @hartnn @Hero @satellite73 @amistre64 @Callisto @experimentX Please help:D

    • one year ago
  2. maheshmeghwal9
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    @lalaly too:)

    • one year ago
  3. hartnn
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    do you know standard formula for 1^2+2^2+3^2+.... n^2 =... ?

    • one year ago
  4. 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??

    • one year ago
  5. maheshmeghwal9
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    no :(

    • one year ago
  6. maheshmeghwal9
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    i mean how to expand?

    • one year ago
  7. ash2326
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    Multiply the terms :) \[(n)(n+1)(2n+1)\]

    • one year ago
  8. maheshmeghwal9
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    ok then i gt this \[3n^2+2n^3+n\]

    • one year ago
  9. ash2326
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    Divide this by 6,

    • one year ago
  10. maheshmeghwal9
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    ok then here it is \[\frac{n^2}{2}+\frac{n^3}{3}+\frac{n}{6}.\]

    • one year ago
  11. ash2326
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    Obviously for positive n \[\large \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} > \frac {n^3}{3} \]

    • one year ago
  12. maheshmeghwal9
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    yeah thanx:)

    • one year ago
  13. ParthKohli
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    How do you conclude that it is “obvious” for a positive \(n\)? @ash2326

    • one year ago
  14. maheshmeghwal9
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    @abhyudaysingh12 got it or not?

    • one year ago
  15. maheshmeghwal9
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    for every 'n' it is true actually:)

    • one year ago
  16. ParthKohli
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    How would you conclude that?

    • one year ago
  17. hartnn
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    \(\large \frac{n^2}{2}+\frac{n}{6} >0\) for n>0

    • one year ago
  18. hartnn
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    ^ that way

    • one year ago
  19. hartnn
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    then add n^3/3 on both sides.

    • one year ago
  20. ParthKohli
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    \[n^2 + n > 0 \iff n(n + 1)>0 \iff n +1>0\iff n>-1\]

    • one year ago
  21. maheshmeghwal9
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    BUT question says that n>0 so n>-1 is ignored 1,2,.........n see @ParthKohli :)

    • one year ago
  22. ParthKohli
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    But you asserted that it's true for ALL \(n\).

    • one year ago
  23. maheshmeghwal9
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    sorry for that statement but that i gt in haste;)

    • one year ago
  24. ParthKohli
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    lol okay

    • one year ago
  25. maheshmeghwal9
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    :D good job;)

    • one year ago
  26. ParthKohli
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    ?

    • one year ago
  27. maheshmeghwal9
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    to discuss something is good job:)

    • one year ago
  28. ParthKohli
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    ?

    • one year ago
  29. maheshmeghwal9
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    ?

    • one year ago
  30. 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.

    • one year ago
  31. UnkleRhaukus
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    |dw:1357174739667:dw|

    • one year ago
  32. maheshmeghwal9
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    |dw:1357869383781:dw|

    • one year ago
  33. 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 \(n\).

    • one year ago
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