$$3\sum\limits_{k=1}^{n} (2k - 3)+\sum\limits_{k=1}^{n} (4 − 5k)$$
Can someone show all steps to simplify this? The answer should simplify down to the sum of $k-5$ I believe.
simplifying a basic sequence
0
$\begingroup$
sequences-and-series
1 Answers
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It is actually very simple: $$3\sum_{k=1}^n (2k-3)+\sum_{k=1}^n (4-5k)$$ Note that: $c\cdot \sum_{k=1}^n (f(k))\equiv \sum_{k=1}^n (c\cdot f(k))$, where $c$ is a constant: $$\sum_{k=1}^n 3(2k-3)+\sum_{k=1}^n (4-5k)=\sum_{k=1}^n (6k-9)+\sum_{k=1}^n (4-5k)$$ Note that the limits of the sums are the same: $$\sum_{k=1}^n (6k-9)+\sum_{k=1}^n (4-5k)=\sum_{k=1}^n (6k-9+4-5k)=\sum_{k=1}^n (k-5)$$