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prove that,
r^m + r^(m+1) + ... + r^(n1) + r^n = (r^(n+1)r^m)/(r1)
 one year ago
 one year ago
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_iitiBest 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_iitiBest ResponseYou've already chosen the best response.0
for prove of GP you can search on google
 one year ago

tanvirzawadBest 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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