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Notasi Sigma

Authored by Riani Widiastuti

Mathematics

KG

CCSS covered

Used 55+ times

Notasi Sigma
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10 questions

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1.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Jumlah yang tepat dari notasi sigma

 i = 25 (2i)\sum_{i\ =\ 2}^5\ \left(2^i\right)  

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2.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Notasi sigma yang tepat untuk menyatakan deret 6 + 13 + 20 + 27 + ... + 48 adalah ....

 i = 16 (7i + 2)\sum_{i\ =\ 1}^6\ \left(7i\ +\ 2\right)  

 i = 17  (7i)\sum_{i\ =\ 1}^7\ \ \left(7i\right)  

 i = 17 (7i  1)\sum_{i\ =\ 1}^7\ \left(7i\ -\ 1\right)  

 i = 17 (7i + 1)\sum_{i\ =\ 1}^7\ \left(7i\ +\ 1\right)  

 i = 16  (7i  2)\sum_{i\ =\ 1}^6\ \ \left(7i\ -\ 2\right)  

3.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

notasi sigma yang tepat dari 3 + 6 +9+ ... + 69 adalah ....

n=0223n\sum_{n=0}^{22}3n

n=1223n\sum_{n=1}^{22}3n

n=1233n\sum_{n=1}^{23}3n

n=0223n\sum_{n=0}^{22}3^n

4.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

 i=14(i23i+2) = .....\sum_{i=1}^4\left(i^{2^{ }}-3i+2\right)\ =\ .....  

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5.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

5. 13 + 23+33+43+...+1031^{3\ }+\ 2^3+3^3+4^3+...+10^3  dapat ditulis  sebagai ...

 n=05n3\sum_{n=0}^5n^3  

 n=110n3\sum_{n=1}^{10}n^3  

 n=1n103\sum_{n=1}^n10^3  

 n=1310n\sum_{n=1}^310^n  

 n=110(1 +n)3\sum_{n=1}^{10}\left(1\ +n\right)^3  

Tags

CCSS.HSF.BF.A.2

6.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

 i=13(i2 +2) i = 25(3i )= .....\sum_{i=1}^3\left(i^2\ +2\right)-\ \sum_{i\ =\ 2}^5\left(3i\ \right)=\ .....  

 i=17 (i2  3i+2)\sum_{i=1}^7\ \left(i^2\ -\ 3i+2\right)  

 i =17 (i2 3i)\sum_{i\ =1}^7\ \left(i^{2\ }-3i\right)  

 i = 17 (i2 3i +4)\sum_{i\ =\ 1}^7\ \left(i^2\ -3i\ +4\right)  

 i =17 (i2 3i +8)\sum_{i\ =1}^7\ \left(i^2\ -3i\ +8\right)  

 i=17 (i2 3i  4)\sum_{i=1}^7\ \left(i_{ }^2\ -3i\ -\ 4\right)  

7.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

 k=512 (2n +1)\sum_{k=5}^{12}\ \left(2n\ +1\right)  

 n=18(2n3)\sum_{n=1}^8\left(2n-3\right) 

 n=18(2n +9)\sum_{n=1}^8\left(2n\ +9\right) 

 n=18(2n7)\sum_{n=1}^8\left(2n-7\right) 

 n=17(2n+9)\sum_{n=1}^7\left(2n+9\right) 

 n=18(2n+5)\sum_{n=1}^8\left(2n+5\right) 

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