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tanvirzawad
Group Title
prove that,
r^m + r^(m+1) + ... + r^(n1) + r^n = (r^(n+1)r^m)/(r1)
 2 years ago
 2 years ago
tanvirzawad Group Title
prove that, r^m + r^(m+1) + ... + r^(n1) + r^n = (r^(n+1)r^m)/(r1)
 2 years ago
 2 years 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 \[a+a*r+a*r ^{2}+ a*r ^{3}+....+a*r ^{n1}=a*\left( r ^{n}  1 \right)/(r1)\]
 2 years ago

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