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panchtatvam
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 ?
|dw:1343194234112:dw|
fundamental theorem of algebra
n=1 is the base case
@experimentX 1 is the base case but it also has to be proved first.
I'm unable to reduce the expression in the second bracket as it involves inverse terms with dont cancel to 1
in your formula n-1 > 0, but n-1 =0 for n=1
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
|dw:1343194693791:dw|
@experimentX is that the final step?
no .. not really, currently, i cannot think using induction.
no it's purely based on induction.....
it proves directly using geometric sum
u prove very dumb things using induction/by contradiction...........
so now by induction.....we have xn−yn=(x−y)(xn−1+xn−2y+.....+xyn−2+yn−1 by induction
nd we need to prove xn+1−yn+1=(x−y)(xn+xny+.....+xyn-1+yn......
so try showing that (x−y)(xn+xny+.....+xyn-1+yn...... = xn+1−yn+1
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.
|dw:1343195660262:dw| i guess ... certainly, other ways are more intuitive
\[ (x-y)(x^n + x^{n-1}y + x^{n-2}y^2 + ... +y^n) \\ = x^n(x-y) + y (x-y)(x^{n-1}+x^{n-2}y + ...+y^{n-2}) \\ = x^n(x-y)+y (x^n - y^n) = x^{n+1} - y^{n+1}\]
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.