I've added an image of how I've approached this problem. Any clarity would be appreciated.
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1$\begingroup$ (3) is wrong and the given answer is also wrong. $\endgroup$– geetha290krmCommented Dec 27, 2022 at 11:44
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1$\begingroup$ Look closely at the index variable in (3). It's not actually a triangle sum! $\endgroup$– Sammy BlackCommented Dec 27, 2022 at 12:07
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1$\begingroup$ You have mistaken $\sum_{i=1}^{n-1} n=\overbrace{n+n+\dots+n}^{n-1 \text{ times}}$ with $\sum_{i=1}^{n-1} i=1+2+\dots+(n-1)$ $\endgroup$– SilCommented Dec 27, 2022 at 12:20
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1 Answer
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Although the triangle number sum is $$ \sum_{i=1}^{n-1} i = \tfrac12 n (n-1), $$ the sum in (3) has the same constant $n$ in each term (the index variable is $i$), hence $$ \sum_{i=1}^{n-1} n = n \sum_{i=1}^{n-1} 1 = n (n-1). $$ Thus, the total (original expression) is $$ n (n-1) + n^2 = 2n^2 - n. $$
answered Dec 27, 2022 at 12:06
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$\begingroup$ In your step 2, it is still unclear to me why you have chosen to take n out of the summation. If not a triangle sum, what is it? $\endgroup$ Commented Dec 27, 2022 at 12:12
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1$\begingroup$ Look closely. The index $i$ runs from $1$ up to $n-1$, but the terms being summed don't depend on $i$, i.e. they are constant. There are $n-1$ terms all the same, so $$n + n + \cdots + n = n (1 + 1 + \cdots + 1) = n (n - 1) $$ $\endgroup$ Commented Dec 27, 2022 at 12:20
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$\begingroup$ Can you share any resources, strategies, or textbooks that have helped you to better understand summation expansion? $\endgroup$ Commented Dec 27, 2022 at 15:15
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$\begingroup$ The main thing is to just write out the terms carefully, treating the expression being summed as a function of the index variable. Sometimes it helps to write out a few terms, i.e. for any $f$ (which may not even explicitly reference $k$), $$ \sum_{k=1}^5 f(k) = f(1) + f(2) + f(3) + f(4) + f(5). $$ Also, keep in mind that the sum on the right-hand side has many different expressions in the summation notation, i.e. we could shift the index and write $$ \sum_{y=2023}^{2027} f(y-2022) $$ $\endgroup$ Commented Dec 27, 2022 at 22:48
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$\begingroup$ If you like, think of generating a table of pairs $(\text{index}, \text{value})$ = $(k, f(k))$ like you might with any function, allowing indices in the range $n \leq k \leq N$: $$ \begin{array}{c|c} k & f(k) \\ \hline n & f(n) \\ n+1 & f(n+1) \\ \vdots & \vdots \\ N & f(N) \end{array} $$ Then, $$ \sum_{k=n}^N f(k) $$ is just the sum of the values in the table! $\endgroup$ Commented Dec 27, 2022 at 22:53