Prove the following statment $\frac{S_m-S_n}{S_{m+n}}=\frac{m-n}{m+n}$ if $S_m$, $S_n$ and $S_{m+n}$ are arthmetic series

Question

I proved this statment using formulas for n-th term of aritmetic sequence and for arithmetic series.

Does anyone know any other proof?

Answer 1

I first used formula for arithmetic series

$\frac{S_m-S_n}{S_{m+n}}=\frac{\frac{m}{2}(a_1+a_m)-\frac{n}{2}(a_1+a_n)}{\frac{m+n}{2}(a_1+a_{m+n})}=\frac{m(a_1+a_m)-n(a_1+a_n)}{(m+n)(a_1+a_{m+n})}=\frac{m(a_1+a_1+(m-1)d)-n(a_1+a_1+(n-1)d)}{(m+n)(a_1+a_1+(m+n-1)d)}=\frac{2ma_1+m(m-1)d-2na_1-n(n-1)d}{(m+n)(2a_1+(m+n-1)d)}$

$=\frac{2a_1(m-n)+[m(m-1)-n(n-1)]d}{(m+n)(2a_1+(m+n-1)d)}=\frac{(m-n)[2a_1+\frac{m(m-1)-n(n-1)]}{m-n}d]}{(m+n)(2a_1+(m+n-1)d)}$ (*)

The desired expression will be given if

$\frac{m(m-1)-n(n-1)}{m-n}=(m+n-1)$

$\frac{-m+n+m^2-n^2}{m-n}=(m+n-1)$

$-1+\frac{(m-n)(m+n)}{m-n}=(m+n-1)$

$-1+(m+n)=(m+n-1)$

Now we can cancel common factors in (*) and starting statment is proven.

Answer 2

Assuming that $ \ S_{m} \ , \ S_{n} \ , \ $ and $ \ S_{m+n} \ \ $ all refer to the same arithmetic progression (the wording of the title is not specific) and that these represent the sum $ \ S_k \ = \ k·a \ + \ \frac{k·(k-1)}{2}·d \ $ for the sequence with terms $ \ a \ + \ (i - 1)·d \ \ , \ \ i \ = \ 1 \ , \ , \ \ldots \ , \ k \ \ , $ then we can alternatively write the sum as $ \ S_k \ = \ \frac{d}{2}·k^2 \ + \ k·\left(a - \frac{d}{2} \right) \ \ . $ This may be considered as a quadratic polynomial with variable $ \ k \ $ with coefficients $ \ D \ = \ \frac{d}{2} \ $ and $ \ A \ = \ \left(a - \frac{d}{2} \right) \ \ , $ which will be specific constants for a given arithmetic progression.

The ratio specified is then $$ \frac{S_m \ - \ S_n}{S_{m+n}} \ \ = \ \ \frac{[ \ D·m^2 \ + \ A·m \ ] \ - \ [ \ D·n^2 \ + \ A·n \ ]}{ D·(m \ + \ n)^2 \ + \ A·(m \ + \ n) } $$ $$ = \ \ \frac{ D·(m^2 \ - \ n^2) \ + \ A·(m \ - \ n)}{ D·(m \ + \ n)^2 \ + \ A·(m \ + \ n) } \ \ = \ \ \frac{ D·(m \ - \ n)·(m \ + \ n) \ + \ A·(m \ - \ n)}{ D·(m \ + \ n)^2 \ + \ A·(m \ + \ n) } $$ $$ = \ \ \frac{ (m \ - \ n) \ · \ [ \ D·(m \ + \ n) \ + \ A \ ]}{ (m \ + \ n) \ · \ [ \ D·(m \ + \ n) \ + \ A \ ] } \ \ = \ \ \frac{ m \ - \ n }{ m \ + \ n } \ \ . $$

We see from this that the same sort of relationship would hold for any quadratic polynomial $ \ P_k \ = \ A·k^2 \ + \ B·k \ + \ C \ \ . $