while n , k , m are integers :(n + (n+1) + (n+2) .... + (n + k))/ ((n + k + 1) + (n + k + 2) ..... + (n + k + m) = (n + (n + k)) / ((n + k + 1) + (n + k + m ) CAN YOU PROVE THIS ? (NOTE: I AM NOT SURE THIS IS ALREADY PROVED OR NOT)
ZorméarMar 18, 2019Hi therelet's sayA = n+(n+1)+...+(n+k) = (n+(n+k))(k+1)/2 B = (n+k+1)+...+(n+k+m) = (n+k+1+n+k+m)(m)/2 (these are just basic sum formulas, arithmetic series)The left hand side is just A/B, so the final result is (n+(n+k))/((n+k+1)+(n+k+m)) * (k+1)/mI think
Hi there
let's say
A = n+(n+1)+...+(n+k) = (n+(n+k))(k+1)/2
B = (n+k+1)+...+(n+k+m) = (n+k+1+n+k+m)(m)/2 (these are just basic sum formulas, arithmetic series)
The left hand side is just A/B, so the final result is (n+(n+k))/((n+k+1)+(n+k+m)) * (k+1)/m
I think
@Zorméar Yes