tolong di jawab dengan cara penyelesaiannya! Nomor 9

Dengan menggunakan jumlah dengan notasi sigma:
[tex]$\begin{align}\sum_{i=1}^n1&=n \\ \sum_{i=1}^nn&=\frac{n(n+1)}{2} \\ \sum_{i=1}^nn^2&=\frac{n(n+1)(2n+1)}{6}\end{align}[/tex]

Diperoleh:
[tex]$\begin{align}&\sum_{a=7}^{12}(a^2+4)-\sum_{a=1}^6(a^2+4a-5) \\ &=\sum_{a=7}^{12}a^2+4\sum_{a=7}^{12}1-\sum_{a=1}^6a^2-4\sum_{a=1}^6a+5\sum_{a=1}^61 \\ &=\left(\sum_{a=1}^{12}a^2-\sum_{a=1}^{6}a^2\right)+4\left(\sum_{a=1}^{12}1-\sum_{i=1}^61\right)-\sum_{a=1}^6a^2-4\sum_{a=1}^6a+5\sum_{a=1}^61 \\ &=\left(\frac{12(12+1)(2.12+1)}{6}-\frac{6(6+1)(2.6+1)}{6}\right)+4(12-6) \\&~~~~-\frac{6(6+1)(2.6+1)}{6}-4\times\frac{6(6+1)}{2}+5\times6\\&=2.13.25-7.13+24-7.13-12.7+30 \\ &=650-91+24-91-84+30\\&=438\end{align}[/tex]