设数列\(\left\{ {{a_n}} \right\}\)前\(n\)项和为\({S_n}\)，若\({a_1} = 1\)，\(2S_n^2 - 2{S_n}{a_n} + {a_n} = 0\left( {n \geqslant 2{\rm{,}}n \in {{\rm{N}}^*}} \right)\)，则\(\sum\limits_{i = 1}^n {\frac{1}{{{S_i}}} = } \)___.
解：\(\because \)当\(n⩾2\)时\(2S_n^2 - 2{S_n}{a_n} + {a_n} = 0\)，\(\therefore {{a}_{n}}=\frac {2{S}^{2}_{n}} {2{{S}_{n}}-1}\)，
\(\therefore {S_n} - {S_{n - 1}} = \frac{{2S_n^2}}{{2{S_n} - 1}}\)，整理可得\({S_n} - {S_{n - 1}} = - 2{S_n}{S_{n - 1}}\)，
\(\therefore \frac{1}{{{S_n}}} - \frac{1}{{{S_{n - 1}}}} = 2\)，数列\(\left\{ {\frac{1}{{{S_n}}}} \right\}\)是以1为首项，2为公差的等差数列，\(\therefore \frac{1}{{{S_n}}} = 2n - 1\)，
\(\therefore \sum\limits_{i = 1}^n {\frac{1}{{{S_i}}} = \frac{{\left( {1 + 2n - 1} \right)n}}{2}} = {n^2}\).
故答案为：\({n^2}\)
