Bank Soal Matematika SMA Notasi Sigma

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Notasi sigma yang ekuivalen dengan $\sum_{k=5}^9\left(2k-4\right)^2$ adalah....

$4\sum_{k=1}^5k^2+16\ \sum_{k=1}^5\left(k+1\right)$

$4\sum_{k=1}^5k^2+\ \sum_{k=1}^5\left(k+1\right)$

$\sum_{k=1}^5k^2+16\ \sum_{k=1}^5\left(k+1\right)$

$4\sum_{k=1}^5k^{ }+16\ \sum_{k=1}^5\left(k+1\right)$

$4\sum_{k=1}^5k^2+16\ \sum_{k=1}^5\left(2k+1\right)$

Menggunakan sifat operasi sumasi

$\sum_{i=k}^nf\left(i\right)=\sum_{i=k+a}^{n+a}f\left(i-a\right)$

maka

$\sum_{k=5}^9\left(2k-4\right)^2=\sum_{k=5-4}^{9-4}\left(2\left(k+4\right)-4\right)^2$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{k=1}^5\left(2k+4\right)^2$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{k=1}^5\left(4k^2+16k+16\right)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\sum_{k=1}^5k^2+16\ \sum_{k=1}^5\left(k+1\right)$

Jadi, notasi sigma yang ekuivalen dengan

$\sum_{k=5}^9\left(2k-4\right)^2=4\sum_{k=1}^5k^2+16\ \sum_{k=1}^5\left(k+1\right)$

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