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A question I've come across says that $\sum_{n=1}^{10} a = 50$. Okay, so far so good. Then it asks me to find $\sum_{n=1}^{10} (4a + 3)$. I looked at the answer and found it was 230, but I just can't figure out why. My best guess would be 203, but that's not the case. Any thoughts?

Caleb Bertrand
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  • $\sum (4a + 3) = \sum 4a + \sum 3 = 4\sum a + \sum 3$ – Doug M Oct 05 '17 at 19:31
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    Notice that $\sum\limits_{n=1}^{10}a=\underbrace{a+a+a+\dots+a}{10~\text{times}}=10a$. Similarly $\sum\limits{n=1}^{10}(4a+3)=\underbrace{(4a+3)+(4a+3)+(4a+3)+\dots+(4a+3)}_{10~\text{times}}$ – JMoravitz Oct 05 '17 at 19:32
  • Oh, I see your point. That means that a must be 5, which means the second equation must be equal to 230. Thanks so much. Sometimes I just need someone else to tell me XD – Caleb Bertrand Oct 05 '17 at 19:38

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$\sum_{n=1}^{10} (4a + 3) = 4\sum_{n=1}^{10} a + \sum_{n=1}^{10} 3 = 4(50) + 3+3+...+3 $

where $3+3+...+3$ is the sum of 10 3's, one for each n.

Thus $\sum_{n=1}^{10} (4a + 3) = 230.$