Similar sums (corrected)

From the given equation it can be clearly seen that $10 s_1 + s_2=n$ .Now I will only use this equation to find the required expression.

$10 s_1+ s_2=n \\ =>9 s_1 +s_1+s_2=n \\ =>9s_1+s_2+s_1 =n-2 s_1 \\=>s_2- s_1=n -11 s_1..............................Eqn(1)$

$10 s_1 +s_2=n \\ =s_1+s_2=n-9 s_1...........................................................Eqn(2)$

Come to the given expression $\dfrac{2n+9(s_2-s_1)}{s_2+s_1} \\ =\dfrac{2n+9(n-11 s_1)}{n-9 s_1} \\ =\dfrac{11n-99 s_!}{n-9 s_1} \\=\dfrac{11(n-9 s_1)}{n-9 s_1}=11$

Let’s assume that for a particular $n$

$\dfrac { 2n+9\left( { S }_{ 2 }-{ S }_{ 1 } \right) }{ { S }_{ 2 }+{ S }_{ 1 } } =11$

so that

${ S }_{ 2 }=n-10{ S }_{ 1 }$

We wish to find the value for the same expression for $n+1$ , which is

$\dfrac { 2\left( n+1 \right) +9\left( { S }_{ 2 }+{ t }_{ 2 }-{ S }_{ 1 }-{ t }_{ 1 } \right) }{ { S }_{ 2 }+{ t }_{ 2 }+{ S }_{ 1 }+{ t }_{ 2 } }$

Where $t_1$ and $t_2$ are the next terms in the series

${ t }_{ 1 }=\left( \dfrac { 9 }{ { 10 }^{ 2\left( n+1 \right) }-1 } \right) \left( \dfrac { { { 10 }^{ 2\left( n+1 \right) -1 }-1 } }{ 9 } \right)$

${ t }_{ 2 }=\left( \dfrac { 9 }{ { 10 }^{ 2\left( n+1 \right) }-1 } \right)$

This expression reduces to $11$ , with $S_1$ and $n$ both dropping out

Thus, since in the case of $n=1$ the first expression has the value of $11$ , by induction it’s true for any value of $n$