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anonymous
 4 years ago
prove using mathematical induction
\[x ^{n}  y^{n} = (xy)(x^{n1} + x^{n2}y +.....+ xy^{n2} + y^{n1}\]
I'm not even able to prove it true for n =1 .
How could one reduce the term in the second bracket to 1 ?
anonymous
 4 years ago
prove using mathematical induction \[x ^{n}  y^{n} = (xy)(x^{n1} + x^{n2}y +.....+ xy^{n2} + y^{n1}\] I'm not even able to prove it true for n =1 . How could one reduce the term in the second bracket to 1 ?

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experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1343194234112:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0fundamental theorem of algebra

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2n=1 is the base case

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@experimentX 1 is the base case but it also has to be proved first.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm unable to reduce the expression in the second bracket as it involves inverse terms with dont cancel to 1

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2in your formula n1 > 0, but n1 =0 for n=1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well u can do it this way.......... (xy)(x^n+y^n+.....) (x^(n)  y^(n))(xy)+ y(xy)(xn−1+xn−2y+.....+xyn−2

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1343194693791:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@experimentX is that the final step?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2no .. not really, currently, i cannot think using induction.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no it's purely based on induction.....

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2it proves directly using geometric sum

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u prove very dumb things using induction/by contradiction...........

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so now by induction.....we have xn−yn=(x−y)(xn−1+xn−2y+.....+xyn−2+yn−1 by induction

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nd we need to prove xn+1−yn+1=(x−y)(xn+xny+.....+xyn1+yn......

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so try showing that (x−y)(xn+xny+.....+xyn1+yn...... = xn+1−yn+1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if 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.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1343195660262:dw i guess ... certainly, other ways are more intuitive

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2\[ (xy)(x^n + x^{n1}y + x^{n2}y^2 + ... +y^n) \\ = x^n(xy) + y (xy)(x^{n1}+x^{n2}y + ...+y^{n2}) \\ = x^n(xy)+y (x^n  y^n) = x^{n+1}  y^{n+1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.
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