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tanvirzawad
Group Title
prove that,
r^m + r^(m+1) + ... + r^(n1) + r^n = (r^(n+1)r^m)/(r1)
 one year ago
 one year ago
tanvirzawad Group Title
prove that, r^m + r^(m+1) + ... + r^(n1) + r^n = (r^(n+1)r^m)/(r1)
 one year ago
 one year ago

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devendra_iiti Group TitleBest ResponseYou've already chosen the best response.0
its a geometric progression(GP). GP has a formula \[a+a*r+a*r ^{2}+ a*r ^{3}+....+a*r ^{n1}=a*\left( r ^{n}  1 \right)/(r1)\]
 one year ago

devendra_iiti Group TitleBest ResponseYou've already chosen the best response.0
for prove of GP you can search on google
 one year ago

tanvirzawad Group TitleBest ResponseYou've already chosen the best response.0
Thank 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]
 one year ago
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