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For this question, I need to express these two sums as a single sum. Here is what I have so far. I'm not sure if I'm doing it right, can anyone please help me?

$$ \Biggl(3\sum \limits_{k=1}^{n} (k^2-1)\Biggl) + \Biggl(5\sum\limits_{k=n+1}^{2n} (k-n)^2+1\Biggl)$$

$$ = \Biggl(3\sum \limits_{k=n+1}^{2n} (k^2-1)+n\Biggl) + \Biggl(5\sum\limits_{k=n+1}^{2n} (k-n)^2+1\Biggl)$$

$$ = \Biggl(\sum \limits_{k=n+1}^{2n} 3(k^2-1)+n + 5(k-n)^2+1\Biggl)$$

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\begin{align} \left(3\sum \limits_{k=1}^{n} (k^2-1)\right) + \left(5\sum\limits_{k=n+1}^{2n} (k-n)^2+1\right)&=\left(3\sum \limits_{k=1}^{n} k^2-1\right) + \left(5\sum\limits_{k=1}^{n} k^2+1\right) \\ &=3\left(\sum \limits_{k=1}^{n} k^2\right)-3n + 5\left(\sum\limits_{k=1}^{n} k^2\right)+5n \\ &=\left(8\sum \limits_{k=1}^{n} k^2\right)+2n \end{align}

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  • $\begingroup$ How did you change the the lower limit from n+1 to 1 in the second sum? $\endgroup$ – user325 Jan 19 '20 at 20:20
  • $\begingroup$ See the answer from @PythonSage, he explains it very well $\endgroup$ – Norse Jan 19 '20 at 20:22
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$\left(3\sum\limits_{1}^{n}(k^2-1)\right)+\left(5\sum\limits_{n+1}^{2n}(k-n)^2+1\right)$ Note that for $k=\{n+1,n+2,\cdots2n\}, k-n=\{1,2,\cdots n\}$. Hence, the sum can be rewritten as $\left(3\sum\limits_{1}^{n}(k^2-1)\right)+\left(5\sum\limits_{1}^{n}(j)^2+1\right)=8\sum\limits_{1}^{n}k^2+2n$.

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  • $\begingroup$ How did you change the lower limit to 1 in the second sum? you just did j = k-n? $\endgroup$ – user325 Jan 19 '20 at 20:24
  • $\begingroup$ The +1 is also included in the summation $\endgroup$ – user325 Jan 19 '20 at 20:26
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    $\begingroup$ $j$ varies from $1$ to $n$ as well. The key idea is that a variable varies from $1$ to $n$. You can write it as $j,k,x,y,z$ right? Hence, we can write both $j$ and $k$ as $k$. Hope that clears your confusion $\endgroup$ – PythonSage Jan 19 '20 at 20:26
  • $\begingroup$ @user325 Edited. $\endgroup$ – PythonSage Jan 19 '20 at 20:27
  • $\begingroup$ See here for info on changing the index of summation if you are unfamiliar with it $\endgroup$ – Norse Jan 19 '20 at 20:28
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Try this: $$\sum_{k = 1}^n 3(k - 1)^2 + \sum_{k = n + 1}^{2n} 5(k - n)^2 + 1$$ $$ = \sum_{k = 1}^n 3(k - 1)^2 + \sum_{k = 1}^{n} 5k^2 + 1$$ $$ = 1 + \frac{3n(n-1)(2n-1)}{6} + \frac{5n(n+1)(2n+1)}{6}$$ $$ = \frac{6 + n(3(2n^2 - 3n + 1) + 5(2n^2+3n+1))}{6}$$ $$ = \frac{8n^3+3n^2+4n + 3}{3}$$

Using the common sum $$\sum_{k = 1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$

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