ધારો કે $S_1$ પહેલા $n$ યુગ્મ પ્રાકૃતિક સંખ્યાઓનો સરવાળો દર્શાવે છે.

$S_1 = 2 + 4 + 6 + ….n$પદો સુધી

$ = \frac{{\text{n}}}{{\text{2}}}{\text{[2}} \times {\text{2}} + {\text{(n}} – {\text{1)}} \times {\text{2]}}\,{\text{[}}\because \,\,{{\text{a}}_{\text{1}}} = {\text{2,}}\,\,{\text{d}} = {\text{2]}}$

$ = \frac{{\text{n}}}{{\text{2}}}{\text{(2n}} + {\text{2)}} = {\text{n(n}} + {\text{1)}}\,\,…..{\text{(1)}}$

$\therefore \,\,{{\text{S}}_{\text{1}}} = \frac{{\text{n}}}{{\text{2}}}{\text{(4}} + {\text{2n}} – {\text{2)}}$

ધારો કે $S_2$ પહેલા $n$ અયુગ્મ પ્રાકૃતિક સંખ્યાઓનો સરવાળો દર્શાવે

$\therefore \,\,\,{{\text{S}}_{\text{2}}} = {\text{1}} + {\text{3}} + {\text{5}} + …..{\text{n }}$ પદો સુધી 

$ = \frac{{\text{n}}}{{\text{2}}}[2 \times 1 + {\text{(n}} – {\text{1)2]}}\,$

$\,{\text{[}}\because \,\,{{\text{a}}_{\text{1}}} = {\text{1,}}\,\,{\text{d}} = {\text{2]}}$

$\therefore \,\,\,{{\text{S}}_{\text{2}}}\,\, = \frac{{\text{n}}}{{\text{2}}}{\text{(2}} + {\text{2n}} – {\text{2)}} = \frac{{\text{n}}}{{\text{2}}}{\text{(2n)}} = {{\text{n}}^{\text{2}}}\,\,……{\text{(2)}}$

સમીકરણ (1) ને સમીકરણ (2), વડે ભાગતાં

$\frac{{{{\text{S}}_{\text{1}}}}}{{{{\text{S}}_{\text{2}}}}} = \frac{{{\text{n(n}} + {\text{1)}}}}{{{{\text{n}}^{\text{2}}}}} = \frac{{{\text{n}} + {\text{1}}}}{{\text{n}}} = {\text{1}} + \frac{{\text{1}}}{{\text{n}}}$

આમ,${{\text{S}}_{\text{1}}} = \left( {{\text{1}} + \frac{{\text{1}}}{{\text{n}}}} \right)\,{{\text{S}}_{\text{2}}}\,$

$\therefore \,\,{\text{k}} = \frac{{{\text{n}} + {\text{1}}}}{{\text{n}}}$