Question

prove that,

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

Answer 1

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

Answer 2

for prove of GP you can search on google

Answer 3

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]