Exercise Zone : Notasi Sigma

Berikut ini adalah kumpulan soal mengenai notasi sigma tipe standar. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN HOTS

No. 1

Diketahui {\displaystyle\sum_{n=1}^{50}(n+2)=}...

\(\begin{aligned} \displaystyle\sum_{n=1}^{50}(n+2)&=\displaystyle\sum_{n=1}^{50}n+\displaystyle\sum_{n=1}^{50}2\\ &=\dfrac{50(50+1)}2+2\cdot50\\ &=1275+100\\ &=1375 \end{aligned}\)

Jadi, {\displaystyle\sum_{n=1}^{50}(n+2)=1375}.

No. 2

Tentukan hasil dari \displaystyle\sum_{k=1}^2k^2+3k+4+\displaystyle\sum_{k=3}^4k^2+3k+4

Grup Telegram

\(\eqalign{ \displaystyle\sum_{k=1}^2k^2+3k+4+\displaystyle\sum_{k=3}^4k^2+3k+4&=\displaystyle\sum_{k=1}^5k^2+3k+4\\ &=\displaystyle\sum_{k=1}^5k^2+\displaystyle\sum_{k=1}^53k+\displaystyle\sum_{k=1}^54\\ &=\dfrac{5(5+1)(2(5)+1)}6+3\dfrac{5(5+1)}2+5\cdot4\\ &=\dfrac{5(6)(11)}6+3\cdot\dfrac{5(6)}2+20\\ &=55+45+20\\ &=\boxed{\boxed{120}} }\)

Jadi, \displaystyle\sum_{k=1}^2k^2+3k+4+\displaystyle\sum_{k=3}^4k^2+3k+4=120.

No. 3

Notasi sigma yang equivalen dengan \displaystyle\sum_{k=1}^{n}k^2+\displaystyle\sum_{k=4}^{n+3}(2k+1) adalah

1. \displaystyle\sum_{k=1}^{n}\left(k^2+2k+1\right)
2. \displaystyle\sum_{k=1}^{n}\left(k^2+2k+5\right)

3. \displaystyle\sum_{k=1}^{n}\left(k^2+2k+7\right)
4. \displaystyle\sum_{k=1}^{n}\left(k^2+2k+11\right)

Grup Telegram

\(\eqalign{ \displaystyle\sum_{k=1}^{n}k^2+\displaystyle\sum_{k=4}^{n+3}(2k+1)&=\displaystyle\sum_{k=1}^{n}k^2+\displaystyle\sum_{k=4-3}^{n+3-3}\left(2(k+3)+1\right)\\ &=\displaystyle\sum_{k=1}^{n}k^2+\displaystyle\sum_{k=1}^{n}\left(2k+6+1\right)\\ &=\displaystyle\sum_{k=1}^{n}k^2+\displaystyle\sum_{k=1}^{n}\left(2k+7\right)\\ &=\boxed{\boxed{\displaystyle\sum_{k=1}^{n}\left(k^2+2k+7\right)}} }\)

Jadi, \displaystyle\sum_{k=1}^{n}k^2+\displaystyle\sum_{k=4}^{n+3}(2k+1)=\displaystyle\sum_{k=1}^{n}\left(k^2+2k+7\right).
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