Question

If $\frac{p_n}{q_n}$ be the $n^{th}$ convergent to $\sqrt{a^2+1}$, show that $\frac{p_2^2+p_3^2+....+p_{n+1}^2}{q_2^2+q_3^2+....+q_{n+1}^2}=\frac{p_n{+}_1{p}_{n+2}-p_1{p}_2}{q_n{+}_1{q}_{n+2}-q_1{q}_2}$.

Answer 1

$\sqrt{a^2+1}=a+\frac{1}{2a+}\frac{1}{2a+}\frac{1}{2a+}......$

$p_{n+1}=2a{p}_{n+1}+p_n$

$p_{n+1}=2a{p}_n+p_{n-1}$

$p_n=2a{p}_n+p_{n-1}$

$p_3=2a{p}_2+p_1$

Multiply these terms with $p_{n+1},p_n,.....p_2$ respectively, add and erase the terms

$p_{n+1}{p}_n,p_n{p}_{n-1},.......p_3{p}_2$ from each side of the sum; we obtain

$p_{n+2}.p_{n+1}=2a(p_{n+1}^2+p_n^2+....p_2^2)+p_1{p}_2$

$p_{n+2}.p_{n+1}-p_1{p}_2=2a(p_{n+1}^2+p_n^2+....p_2^2)$

$q_{n+2}=2a{q}_{n+1}+q_n$

$q_{n+1}=2a{q}_n+q_{n-1}$

$q_n=2a{q}_{n-1}+q_{n-2}$

$q_3=2a{q}_2+q_1$

Multiply these equations by $q_{n+1},q_n,.....q_2$ respectively, add and erase the terms,

$q_{n+2}.q_{n+1}=2a(q_{n+1}^2+q_n^2+....+q_2^2)+q_1{q}_2$

$q_{n+2}.q_{n+1}-q_1{q}_2=2a(q_{n+1}^2+q_n^2+....+q_2^2)$

$\frac{p_{n+1}.p_{n+2}-p_1{p}_2}{q_{n+1}.q_{n+2}-q_1{q}_2}=\frac{p_2^2+p_3^2+......+p_{n+1}^2}{q_2^2+q_3^2+......+q_{n+1}^2}$