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maheshmeghwal9

  • one year ago

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

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  1. maheshmeghwal9
    • one year ago
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    @UnkleRhaukus @hartnn @Hero @satellite73 @amistre64 @Callisto @experimentX Please help:D

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

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

  4. ash2326
    • one year ago
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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??

  5. maheshmeghwal9
    • one year ago
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    no :(

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

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

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

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

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

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

  12. maheshmeghwal9
    • one year ago
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    yeah thanx:)

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

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

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

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

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

  18. hartnn
    • one year ago
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    ^ that way

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

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

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

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

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

  24. ParthKohli
    • one year ago
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    lol okay

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

  26. ParthKohli
    • one year ago
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    ?

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

  28. ParthKohli
    • one year ago
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    ?

  29. maheshmeghwal9
    • one year ago
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    ?

  30. experimentX
    • one year ago
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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.

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

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

  33. klimenkov
    • one year ago
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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 \(n\).

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