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Mathmuse
 2 years ago
Best ResponseYou've already chosen the best response.0Bring the (r1) to the other side and you would get: \[\huge{(r1)(r^m+r^{m+1}+...+r^{n1}+r^n)=r^{n+1}r^m}\] expand the left side and watch stuff disappear: \[\huge{r^{m+1}r^m+r^{m+2}r^{m+1}+...}\] \[\huge{...+r^{n+11}r^{n1}+r^{n+1}r^n}\] like so: \[\huge{\cancel{r^{m+1}}r^m+r^{m+2}\cancel{r^{m+1}}+...}\] \[\huge{...+\cancel{r^{n}}r^{n1}+r^{n+1}\cancel{r^n}}\]

tanvirzawad
 2 years ago
Best ResponseYou've already chosen the best response.0Thank you brother! we have to prove that, r^(M) + r^(M+1) + ... + r^(N1) + r^N = (r^(N+1)r^M)/(r1); [r not equal to 1] let, x = r^(M) + r^(M+1) + ... + r^(N1) + r^N rx = r^(M+1) + ... + r^(N1) + r^N + r^(N+1) rx + r^M = r^(M) + r^(M+1) + ... + r^(N1) + r^N + r^(N+1) rx + r^M = x + r^(N+1) rx  x = r^(N+1)  r^M x(r1) = r^(N+1)  r^M x = (r^(N+1)  r^M)/(r1); [r not equal to 1] r^(M) + r^(M+1) + ... + r^(N1) + r^N = (r^(N+1)r^M)/(r1) [proved]
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