Question
The sum of $n$ terms of an arithmetic sequence:$$\begin{aligned}a_{1}+\left(a_{1}+d\right)+\left(a_{1}+2 d\right)+& \cdots+\left[a_{1}+(n-1) d\right] \\&=\frac{n}{2}\left[2 a_{1}+(n-1) d\right]\end{aligned}$$
Step 1
We substitute $n=1$ into the formula to get $a_1 = \frac{1}{2}[2a_1 + (1-1)d] = a_1$. This shows that the formula holds for $n=1$. Show more…
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