(1/27)+(1/29)+(1/32)+(1/35)+(1/38)+(1/45)+(1/43)+(1/40)+(1/37)+(1/34)+(1/32)+(1/52)
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MathBot Answer:
\[ \frac{1}{27} + \frac{1}{29} + \frac{1}{32} + \frac{1}{35} + \frac{1}{38} + \frac{1}{45} + \frac{1}{43} + \frac{1}{40} + \frac{1}{37} + \frac{1}{34} + \frac{1}{32} + \frac{1}{52} = \frac{981479972771}{2929313434320} \approx 0.33505461084222867921701779306781 \]
The sum can be computed by rewriting each term over a common denominator, where the least common denominator of \(\frac{1}{27}\), \(\frac{1}{29}\), \(\frac{1}{32}\), \(\frac{1}{35}\), \(\frac{1}{38}\), \(\frac{1}{45}\), \(\frac{1}{43}\), \(\frac{1}{40}\), \(\frac{1}{37}\), \(\frac{1}{34}\), \(\frac{1}{32}\), and \(\frac{1}{52}\) is \(5858626868640\). \[ \begin{aligned} \frac{1}{27} + \frac{1}{29} + \frac{1}{32} + \frac{1}{35} + \frac{1}{38} + \frac{1}{45} + \frac{1}{43} + \frac{1}{40} + \frac{1}{37} + \frac{1}{34} + \frac{1}{32} + \frac{1}{52} &= \frac{1 \cdot 216986180320}{27 \cdot 216986180320} + \frac{1 \cdot 202021616160}{29 \cdot 202021616160} + \frac{1 \cdot 183082089645}{32 \cdot 183082089645} + \frac{1 \cdot 167389339104}{35 \cdot 167389339104} + \frac{1 \cdot 154174391280}{38 \cdot 154174391280} + \frac{1 \cdot 130191708192}{45 \cdot 130191708192} + \frac{1 \cdot 136247136480}{43 \cdot 136247136480} + \frac{1 \cdot 146465671716}{40 \cdot 146465671716} + \frac{1 \cdot 158341266720}{37 \cdot 158341266720} + \frac{1 \cdot 172312554960}{34 \cdot 172312554960} + \frac{1 \cdot 183082089645}{32 \cdot 183082089645} + \frac{1 \cdot 112665901320}{52 \cdot 112665901320} \\ &= \frac{216986180320 + 202021616160 + 183082089645 + 167389339104 + 154174391280 + 130191708192 + 136247136480 + 146465671716 + 158341266720 + 172312554960 + 183082089645 + 112665901320}{5858626868640} \\ &= \frac{1962959945542}{5858626868640} \\&= \frac{981479972771}{2929313434320} \end{aligned} \]