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find nth partial sum of the arithmetic sequence an= 3n+2 n=10
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\[\sum_{n=0}^{10}(3n+2)\] \[3\sum_{n=0}^{10}n+\sum_{n=0}^{10}2\]
\[3\frac{n(n+1)}{2}+2n\] if i recall correctly
Use the variation on the Gauss formula\[s=\frac{n}{2}\left[ 2a _{1}+(n-1)d \right]\]where a1=3(1)+3=5 is the first term n=10 is the number of terms, and d=3 is the common difference of the arth. series
that might only be from n=1,2,3 ....
my set up that is since n=0 results in 0
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yes, you are correct, my assumption is pos. integers for domain
either way will work \[3\times \frac{10\times 11}{2}+20\]
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