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panchtatvam Group Title

prove using mathematical induction \[x ^{n} - y^{n} = (x-y)(x^{n-1} + x^{n-2}y +.....+ xy^{n-2} + y^{n-1}\] I'm not even able to prove it true for n =1 . How could one reduce the term in the second bracket to 1 ?

  • 2 years ago
  • 2 years ago

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  1. experimentX Group Title
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    |dw:1343194234112:dw|

    • 2 years ago
  2. panlac01 Group Title
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    fundamental theorem of algebra

    • 2 years ago
  3. experimentX Group Title
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    n=1 is the base case

    • 2 years ago
  4. panchtatvam Group Title
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    @experimentX 1 is the base case but it also has to be proved first.

    • 2 years ago
  5. panchtatvam Group Title
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    I'm unable to reduce the expression in the second bracket as it involves inverse terms with dont cancel to 1

    • 2 years ago
  6. experimentX Group Title
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    in your formula n-1 > 0, but n-1 =0 for n=1

    • 2 years ago
  7. A.Avinash_Goutham Group Title
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    well u can do it this way.......... (x-y)(x^n+y^n+.....) (x^(n) - y^(n))(x-y)+ y(x-y)(xn−1+xn−2y+.....+xyn−2

    • 2 years ago
  8. experimentX Group Title
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    |dw:1343194693791:dw|

    • 2 years ago
  9. A.Avinash_Goutham Group Title
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    @experimentX is that the final step?

    • 2 years ago
  10. experimentX Group Title
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    no .. not really, currently, i cannot think using induction.

    • 2 years ago
  11. A.Avinash_Goutham Group Title
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    no it's purely based on induction.....

    • 2 years ago
  12. experimentX Group Title
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    it proves directly using geometric sum

    • 2 years ago
  13. A.Avinash_Goutham Group Title
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    u prove very dumb things using induction/by contradiction...........

    • 2 years ago
  14. A.Avinash_Goutham Group Title
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    so now by induction.....we have xn−yn=(x−y)(xn−1+xn−2y+.....+xyn−2+yn−1 by induction

    • 2 years ago
  15. A.Avinash_Goutham Group Title
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    nd we need to prove xn+1−yn+1=(x−y)(xn+xny+.....+xyn-1+yn......

    • 2 years ago
  16. A.Avinash_Goutham Group Title
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    so try showing that (x−y)(xn+xny+.....+xyn-1+yn...... = xn+1−yn+1

    • 2 years ago
  17. panchtatvam Group Title
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    if we follow induction methods then as per @experimentX the formula is valid only for natural indexes . so n =1 gets proved . for n+1 could be proved by solving the RHS instead of adding any term to the value for the equation for n.

    • 2 years ago
  18. experimentX Group Title
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    |dw:1343195660262:dw| i guess ... certainly, other ways are more intuitive

    • 2 years ago
  19. experimentX Group Title
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    \[ (x-y)(x^n + x^{n-1}y + x^{n-2}y^2 + ... +y^n) \\ = x^n(x-y) + y (x-y)(x^{n-1}+x^{n-2}y + ...+y^{n-2}) \\ = x^n(x-y)+y (x^n - y^n) = x^{n+1} - y^{n+1}\]

    • 2 years ago
  20. panchtatvam Group Title
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    I wanted to have a mathematical Induction proof of the problem . But as it comes out the problem needs to have certain assumptions and can't be explained using mathematical induction in the normal way. Assumptions : 1. n > 1 so as shown by @experimentX we need to work the problem from RHS to LHS to prove the second condition of the induction thoerem.

    • one year ago
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