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I've been shown that : $$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$ Now I need to write an explicit formula for the sum: $$\sum_{i=1}^n {(3i+1)}$$

I've come up with an answer that is: $$\sum_{i=1}^n {(3i+1)} = \frac{9n^2 + 6n + 3}{2}$$

However I am not sure if this is right or not. Thanks in advance for any help.

jksk
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4 Answers4

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$$ \sum_{i=1}^n(3i+1)=3\sum_{i=1}^n i+\sum_{i=1}^n 1=\dots $$

egreg
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$$\sum_{i=1}^{n}(3i+1) = 3\sum_{i=1}^{n}i + \sum_{i=1}^{n}1 =\\ 3\frac{n(n+1)}{2} + n = \frac{3n^2+5n}{2}$$

Anton Grudkin
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There is a trick that sometimes make some problems easy to attack: $$\sum_{i=1}^nf(i)=f(1)+\cdots+f(n)=f(n)+\cdots+f(1)=\sum_{i=1}^nf(n+1-i).$$ In this case $f(i)=3i+1$, so we get $$ \begin{align} \sum_{i=1}^n {(3i+1)}&=\frac{1}{2}\left(\sum_{i=1}^n {(3i+1)}+\sum_{i=1}^n {(3(n+1-i)+1)} \right)\\ &=\frac{1}{2}\sum_{i=1}^n {\left((3i+1)+(3(n+1-i)+1)\right)}\\ &=\frac{1}{2}\sum_{i=1}^n {(3n+5)}\\ &=\frac{n(3n+5)}{2}. \end{align} $$

Woria
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Of course, egreg has outlined the optimal way of obtaining your closed-form expression, but if you are learning about sequences and series, then I imagine the "intended" solution was via the $n$th partial sum for arithmetic sequences, namely $S_n=\frac{n(a+a_n)}{2}$, where $a_n$ is the $n$th term ($a_n=a+(n-1)d$, where $d$ is the common difference) and $a$ the first time. Using the information you have given, it is clear that $a=4$ and $a_n=4+(n-1)3=1+3n$. Hence, $$ S_n=\frac{n[4+(1+3n)]}{2}=\frac{n(5+3n)}{2}=\frac{3n^2+5n}{2}, $$ confirming the "slicker" answers provided by others already.

Daniel W. Farlow
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  • Learning sequences and series in school right now, can confirm this is one of the ways to do this. Another would be to write $S_n = n\frac{2a_1+(n-1)d}{2}$ and rearrange. – Jack Pan Jun 14 '16 at 19:33