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)=}...

$\begin{array}{rl}{\displaystyle \sum _{n=1}^{50}(n+2)}& ={\displaystyle \sum _{n=1}^{50}n+{\displaystyle \sum _{n=1}^{50}2}}\\ & ={\displaystyle \frac{50(50+1)}{2}}+2\cdot 50\\ & =1275+100\\ & =1375\end{array}$

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

$\begin{array}{rl}{\displaystyle \sum _{k=1}^2{k}^2+3k+4+{\displaystyle \sum _{k=3}^4{k}^2+3k+4}}& ={\displaystyle \sum _{k=1}^5{k}^2+3k+4}\\ & ={\displaystyle \sum _{k=1}^5{k}^2+{\displaystyle \sum _{k=1}^{5}3k+{\displaystyle \sum _{k=1}^{5}4}}}\\ & ={\displaystyle \frac{5(5+1)(2(5)+1)}{6}}+3{\displaystyle \frac{5(5+1)}{2}}+5\cdot 4\\ & ={\displaystyle \frac{5(6)(11)}{6}}+3\cdot {\displaystyle \frac{5(6)}{2}}+20\\ & =55+45+20\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 120}}}}\end{array}$

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

$\begin{array}{rl}{\displaystyle \sum _{k=1}^n{k}^2+{\displaystyle \sum _{k=4}^{n+3}(2k+1)}}& ={\displaystyle \sum _{k=1}^n{k}^2+{\displaystyle \sum _{k=4-3}^{n+3-3}(2(k+3)+1)}}\\ & ={\displaystyle \sum _{k=1}^n{k}^2+{\displaystyle \sum _{k=1}^n(2k+6+1)}}\\ & ={\displaystyle \sum _{k=1}^n{k}^2+{\displaystyle \sum _{k=1}^n(2k+7)}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \sum _{k=1}^n(k^2+2k+7)}}}}}\end{array}$

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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