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Hi I cannot for the life of me remember how to use the arithmetic progression formula can someone help?

I just need to find it for this sequence:

$$5n^2 + (5n^2 - 1) + (5n^2 - 2) + \ldots + (5n^2 - n)$$

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    $\begingroup$ do you know what is $\sum_{k=1}^m k$? $\endgroup$
    – Surb
    Dec 8, 2015 at 13:00

3 Answers 3

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The term $5n^2$ appears $(n+1)$ times. (Why).

All that is left to do is add the negative of $1, 2, 3 \cdots n$.

You know the sum for 1 to $n$, and that's it.

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    $\begingroup$ Nice, short and sweet. +1 $\endgroup$
    – Shailesh
    Dec 8, 2015 at 13:07
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We have \begin{align*}5n^2 + (5n^2 - 1) + (5n^2 - 2) + \ldots + (5n^2 - n)&= \overbrace{5n^2+5n^2+\ldots+5n^2}^{(n+1) \text{ times}}-(1+2+\ldots+n)\\ &= 5n^2(n+1)-(1+2+\ldots+n)\end{align*} Now, if $S=1+2+\ldots+n$, then \begin{align*}2S &= (1+2+\ldots+n)+(1+2+\ldots+n)=(n+1)+(n-1+2)+\ldots+\big(n-(n-1)+n\big)\\ &=n(n+1)\end{align*} So that $S=n(n+1)/2$ and thus $$ 5n^2 + (5n^2 - 1) + (5n^2 - 2) + \ldots + (5n^2 - n)= 5n^2(n+1)-\frac{n(n+1)}{2}=\frac{n (n+1) (10 n-1)}{2}$$

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As rules of thumb:
An arithmetic sum is the average value of the terms multiplied by the number of terms.
The average of the terms of an arithmetic sequence is the average of the first and last term.

In formulas; if $\left\{ t_k \right\}_{k=0}^n$ is an arithmetic sequence, and $$ S_n=\sum_{k=0}^n t_k $$ then $$ S_n = n\left(\frac{t_0+t_n}{2}\right) $$

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