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tanvirzawad

  • 3 years ago

prove that, r^m + r^(m+1) + ... + r^(n-1) + r^n = (r^(n+1)-r^m)/(r-1)

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  1. Mathmuse
    • 3 years ago
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    Bring the (r-1) to the other side and you would get: \[\huge{(r-1)(r^m+r^{m+1}+...+r^{n-1}+r^n)=r^{n+1}-r^m}\] expand the left side and watch stuff disappear: \[\huge{r^{m+1}-r^m+r^{m+2}-r^{m+1}+...}\] \[\huge{...+r^{n+1-1}-r^{n-1}+r^{n+1}-r^n}\] like so: \[\huge{\cancel{r^{m+1}}-r^m+r^{m+2}-\cancel{r^{m+1}}+...}\] \[\huge{...+\cancel{r^{n}}-r^{n-1}+r^{n+1}-\cancel{r^n}}\]

  2. tanvirzawad
    • 3 years ago
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    Thank you brother! we have to prove that, r^(M) + r^(M+1) + ... + r^(N-1) + r^N = (r^(N+1)-r^M)/(r-1); [r not equal to 1] let, x = r^(M) + r^(M+1) + ... + r^(N-1) + r^N rx = r^(M+1) + ... + r^(N-1) + r^N + r^(N+1) rx + r^M = r^(M) + r^(M+1) + ... + r^(N-1) + r^N + r^(N+1) rx + r^M = x + r^(N+1) rx - x = r^(N+1) - r^M x(r-1) = r^(N+1) - r^M x = (r^(N+1) - r^M)/(r-1); [r not equal to 1] r^(M) + r^(M+1) + ... + r^(N-1) + r^N = (r^(N+1)-r^M)/(r-1) [proved]

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