If \[\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\]\[\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+=\frac{a}{b}\] and HCF \[\mathbf{(a,}\,\,\mathbf{b)=1}\] then (a + b) = ?

• A. 156
• B. 162
• C. 165
• D. 126