Question

Find the sum or quantity without completely expanding or calculating any sums.

Given$\sum _{k=0}^{25}k=325$and$\sum _{k=3}^{28}{\left(k-3\right)}^2=14,910$, find the value of$\sum _{k=3}^{25}\left(k^2-5k+9\right)$.

Short answer

The sum is$13,502$

Step-by-step solution

Step 1. Given information

$\sum _{k=0}^{25}k=325$

$\sum _{k=3}^{28}{\left(k-3\right)}^2=14,910$

Step 2. Rewrite the sum:

$$\begin{gathered} \sum _{k=3}^{25}\left(k^2-5k+9\right) \\ =\sum _{k=3}^{25}\left(k^2-6k+k+9\right) \\ =\sum _{k=3}^{25}\left(k^2-6k+9+k\right) \\ =\sum _{k=3}^{25}\left({(k-3)}^2+k\right) \\ =\sum _{k=3}^{25}{\left(k-3\right)}^2+\underset{k=3}{\overset{25}{\sum k}} \end{gathered}$$

Step 3. To find ∑k=325k

$$\begin{gathered} \sum _{k=0}^{25}k=325 \\ 0+1+2+\sum _{k=3}^{25}k=325 \\ 3+\underset{k=3}{\overset{25}{\sum k}}=325 \\ \underset{k=3}{\overset{25}{\sum k}}=325-3 \\ \underset{k=3}{\overset{25}{\sum k}}=322 \end{gathered}$$

Step 4. To find ∑k=325k-32

$$\begin{gathered} \sum _{k=3}^{28}{\left(k-3\right)}^2=14,910 \\ \sum _{k=3}^{25}{\left(k-3\right)}^2+{(26-3)}^2+{(27-3)}^2+{(28-3)}^2=14,910 \\ \sum _{k=3}^{25}{\left(k-3\right)}^2+{\left(23\right)}^2+{\left(24\right)}^2+{\left(25\right)}^2=14,910 \\ \sum _{k=3}^{25}{\left(k-3\right)}^2+529+576+625=14,910 \\ \sum _{k=3}^{25}{\left(k-3\right)}^2+1730=14,910 \\ \sum _{k=3}^{25}{\left(k-3\right)}^2=14,910-1730 \\ \sum _{k=3}^{25}{\left(k-3\right)}^2=13,180 \end{gathered}$$

Step 5. Substitute the value of the sum obtained in steps 3 and 4 into step 2.

$$\begin{gathered} \sum _{k=3}^{25}\left(k^2-5k+9\right) \\ =\sum _{k=3}^{25}{\left(k-3\right)}^2+\sum _{k=3}^{25}k \\ =13180+322 \\ =13502 \end{gathered}$$