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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. devendra_iiti
    • 3 years ago
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    its a geometric progression(GP). GP has a formula \[a+a*r+a*r ^{2}+ a*r ^{3}+....+a*r ^{n-1}=a*\left( r ^{n} - 1 \right)/(r-1)\]

  2. devendra_iiti
    • 3 years ago
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    for prove of GP you can search on google

  3. 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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