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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 ?
 one year ago
 one year 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 ?
 one year ago
 one year ago

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experimentXBest ResponseYou've already chosen the best response.2
dw:1343194234112:dw
 one year ago

panlac01Best ResponseYou've already chosen the best response.0
fundamental theorem of algebra
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
n=1 is the base case
 one year ago

panchtatvamBest ResponseYou've already chosen the best response.0
@experimentX 1 is the base case but it also has to be proved first.
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.2
in your formula n1 > 0, but n1 =0 for n=1
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.1
well u can do it this way.......... (xy)(x^n+y^n+.....) (x^(n)  y^(n))(xy)+ y(xy)(xn−1+xn−2y+.....+xyn−2
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1343194693791:dw
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.1
@experimentX is that the final step?
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
no .. not really, currently, i cannot think using induction.
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.1
no it's purely based on induction.....
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
it proves directly using geometric sum
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.1
u prove very dumb things using induction/by contradiction...........
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.1
so now by induction.....we have xn−yn=(x−y)(xn−1+xn−2y+.....+xyn−2+yn−1 by induction
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.1
nd we need to prove xn+1−yn+1=(x−y)(xn+xny+.....+xyn1+yn......
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.1
so try showing that (x−y)(xn+xny+.....+xyn1+yn...... = xn+1−yn+1
 one year ago

panchtatvamBest ResponseYou've already chosen the best response.0
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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1343195660262:dw i guess ... certainly, other ways are more intuitive
 one year ago

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

panchtatvamBest ResponseYou've already chosen the best response.0
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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